Basic Fourier integrals

Peter Haggstromwww.gotohaggstrom.com

mathsatbondibeach@gmail.com

August 3, 2014

1 Introduction

”The series a02 +

∑∞n=1(an cos nx+bn sin nx) converges, and indeed uniformly, if

∑(|an|+ |bn|)

converges. Apart from this trivial case the convergence of trigonometric series is a delicateproblem”. A Zygmund, ”Trigonometric Series”, Volume 1, Cambridge University Press,1959, page 4)

Fourier theory is a profound subject which is like a mathematical version of kikuyu grass withrunners that go in every direction and eventually cover the entire ground. Learning it as a stu-dent only gives you the smallest of insights into the breadth of the subject but one has to startsomewhere. In this paper what I have done is provide the essential building blocks of Fourieranalysis in terms of the integral transform and in so doing I have followed the approach of EliasStein and Rami Shakarchi in their superb Princeton Lecture Series [3]. There is a handful ofbasic properties which, when applied to the study of differential equations in particular, will takeyou a long way. To this end I have set out the basic properties with detailed proofs based uponthe assumption that the functions inhabit Schwartz space. One can do Fourier theory withoutreferring to Schwartz space and indeed this is done in many introductory courses, but ultimatelyyou will need to visit Schwartz space. Indeed, as will be shown shortly, the fact that it is possi-ble to get away without Schwartz space is instructive in itself because the theory still works. Inengineering contexts this sort of detail is generally skated over but for those who want to reallyknow why the theory works as well as it does you have to get your hands dirty with the nittygritty of functional analysis. At one level Fourier theory is schizophrenic - it is possible to teachit with relatively little rigour even though analytical tripwires cover the entire ground, or onecan adopt a highly analytical approach which can obscure the remarkable physical aspects ofthe theory. I have tried to steer a middle course and to do so I chosen a variant of the basic heatequation and the Black-Scholes partial differential equation as mechanisms to demonstrate howthe theory works in detail. To give an idea of the sheer scope of Fourier theory here are someexamples.

Typical applications of classical Fourier analysis are to:

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• Frequency Modulation: Alternating current,radio transmission;• Mathematics: Ordinary and partial differential equations, analysis of linear and nonlinear op-erators;Number Theory;•Medicine: Electrocardiography, magnetic resonance imaging, biological neural systems;• Optics and Fibre-Optic Communications: Lens design, crystallography, image processing;• Radio,Television, Music Recording: Signal compression, signal reproduction, filtering;• Spectral Analysis: Identification of compounds in geology, chemistry, biochemistry, massspectroscopy;• Telecommunications: Transmission and compression of signals, filtering of signals, frequencyencoding.

For a contemporary application of Fourier theory and wavelet theory in the context of analysingclimate change, see mathematical physicist John Baez’s Azimuth Project: http://johncarlosbaez.wordpress.com. His website contains material which explains how Gabor transforms are usedin the context of searching for patterns in the Southern Oscillation Index data. That index isrelevant to El Nino and La Nina phenomena.

The discovery of the double helix structure of DNA is contained in the famous x-ray diffraction”photograph 51” generated by Rosalind Franklin and Maurice Wilkins and they used Fourier the-ory and Bessel functions to define the structure in a form that looked like this: Fn = Jn(2πrR) ein(φ+ π

2 )

where Jn(u) is the nth order Bessel function of u [see [1] ]. James Watson who, along with Fran-cis Crick, shared the Nobel Prize for the discovery of DNA apparently sneaked a look at theX-ray photograph on Franlin’s desk and realised that it implied a double helix structure.

Even more remarkable at one level is the use of Fourier theory in number theory. An impor-tant example is Timothy Gower’s proof of Szemeredi’s Theorem [19] which states the follow-ing:

2

Let k be a positive integer and let δ > 0. There exists a positive integer N = N(k, δ) suchthat every subset of the set 1, 2, . . . . ,N of size at least δN contains an arithmetic progression oflength k.

The proof involves consideration of subsets of ZN where N is prime and for a function f : ZN →

C and r ∈ ZN . We set:

f (r) =∑

s∈ZN f (s)ω−rs where ω = e2πiN . The function f is the discrete Fourier transform of f

and is used widely in analytic number theory. Indeed in his paper [19] Gowers makes the pointthat four fundamental properties of the Fourier transform so defined are used repeatedly. Thusif the convolution is written in non-standard form as:

f ∗ g(s) =∑

t∈ZN f (t) g(t − s)

then the following four fundamental identities hold:

( f ∗ g)(r) = f (r) g(r)∑r f (r) g(r) = N

∑s f (s) g(s)∑

r | f (r)|2 = N∑

s| f (s)|2

f (s) = 1N

∑r f (r)ωrs

The first identity tells us that convolutions transform to pointwise products, the second and thirdare Parseval’s identities and the last is the inversion formula.

Beyond Fourier theory is wavelet analsyis and more. The applications that are being developedfor wavelet analysis are very similar to those just listed. But the wavelet algorithms give rise tofaster and more accurate image compression, faster and more accurate signal compression, andbetter denoising techniques that preserve the original signal more completely. The applicationsin mathematics lead, in many situations, to better and more rapid convergence results.

2 Background

At its most basic level we can guarantee the existence of the Fourier transform and its inversewith a simple test that doesn’t require profound analysis. Thus if

∫ ∞−∞| f (x)| dx < ∞ then F f and

F ∗ f = F −1 f exist and are continuous.

We can establish existence by noting that:

|F f (ξ)| =∣∣∣∣ ∫ ∞−∞ f (x) e−2πixξ dx

∣∣∣∣ ≤ ∫ ∞−∞| f (x)| |e−2πixξ | dx =

∫ ∞−∞| f (x)| dx < ∞

Continuity follows from the following estimate for any ξ and ξ′

:

|F f (ξ)−F f (ξ′

)| =∣∣∣∣ ∫ ∞−∞ f (x) e−2πixξ dx−

∫ ∞−∞

f (x) e−2πixξ′ dx∣∣∣∣ =

∣∣∣∣ ∫ ∞−∞ f (x) (e−2πixξ − e−2πixξ′) dx∣∣∣∣

≤∫ ∞−∞| f (x)| |e−2πixξ − e−2πixξ′ | dx

3

Because∫ ∞−∞| f (x)| dx < ∞ we can take the limit ξ

′

→ ξ so that |e−2πixξ − e−2πixξ′ | → 0 and weget |F f (ξ)−F f (ξ

′

)| → 0 as ξ′

→ ξ. This shows that F f is continuous and we can replicate theargument to show that F ∗ f is continuous. The argument can be made rigorous with the usualε, δ,N paraphenalia and there is plenty of that to come.

The fact that the Fourier transform is continuous is non-trivial. The classical electrical engi-neering function (which reflects a physical reality - just connect an oscilloscope to the appro-priate circuit) of the ”box” function or square wave Π(ξ) which is 1 on the interval [−1

2 ,12 ] and

zero otherwise. It is discontinuous but its Fourier transform is: Π(ξ) =∫ ∞−∞

e−2πixξ Π(x) dx =∫ 12

− 12

e−2πixξ dx = sinc ξ, which is continuous. Note that∫ ∞−∞|Π(x) | dx =

∫ 12

− 12

1 dx = 1 so that

Π ∈ L1(R). Now here’s the problem. The sinc function does not satisfy the simple absoluteintegrability condition mentioned above. In fact,

∫ ∞−∞| sinc x | dx = ∞. If sinc x were absolutely

integrable we could be assured that its Fourier transform - which is Π(ξ) - exists and is con-tinuous, but we know that Π is not continuous. One can establish that sinc x is not absolutelyintegrable by a careful estimation process which is instructive in that it shows that although|sinc ξ| =

∣∣∣∣ sin πξπξ

∣∣∣∣ → 0 as ξ → ±∞, the 1ξ factor does not do so fast enough to get convergence of

the integral.

Now it can be shown with some pretty subtle analysis that∫ ∞−∞

e−2πixξ sinc x dx =

1 if |ξ| < 12

0 if |ξ| > 12

Note that you have two oscillatory processes going on in the integrand and they operate in away to achieve cancellations and this is in itself a subtle process that is indicative of the field. Itgets worse. Try to find the Fourier transform of something as basic as cos 2πx using the ideasdeveloped above and you won’t succeed notwithstanding the fact that a cosine signal or a sinesignal underpins western civilisation! This problem is ultmately resolved with the theory oftempered distributions and you get 1

2δ(ξ−1)+ 12δ(ξ+1) where δ(x) is the Dirac delta ”function”.

Indeed, if you pick up a book on spectroscopy, for instance, you will find something like thefollowing (see [21]):

δ(t) =∫ ∞−∞

e+2πits ds = F (1) =∫ ∞−∞

cos(2πts) ds

Suffice it to say that the world of spectroscopy has not crumbled under the weight of suchunrigorous formalism, yet the full justification for such matters does require quite a bit of workon generalised functions which is not the purpose of this paper. To read more on the details ofgeneralised functions see [22].

In short the full theory of tempered distributions (see [20] or more specialised textbooks formore details) provides a rigorous foundation for all the ”usual suspects” of the real world ofelectrical engineering and signal analysis and much more. What follows is a basic introductionto the characteristics of the Schwartz space to show the power of the concepts of ”tempereddistributions” and ”generalised functions”.

The fact that the e−πx2is its own Fourier transform is one of the most fundamental and in-

teresting facts of Fourier theory. The crux of the Heisenberg Uncertainty Principle is that if

4

something is ”localised” in one space it is ”spread out” in another. In other words, in somegeneral sense a function and its Fourier transform cannot be fundamentally localised. More-over, the Gaussian e−πx2

(subject to some scaling) causes the product of momentum uncer-tainty and position uncertainty ∆p ∆q to be minimised. [see [1] page 19 ]. The Fourier trans-form is a bijective mapping on the space of Schwartz functions (for a proof see [2] pages140-142). Thus the Gaussian e−α x2

, where α > 0, inhabits Schwartz space and is a fixedpoint of the space. If F represents the process of taking a Fourier transform of a function,F (e−πx2

) = e−πξ2

where F ( f )(x) = f (ξ) =∫ ∞−∞

f (x) e−2πixξ dx while the inverse transform isgiven by F ∗( f )(ξ) =

∫ ∞−∞

f (ξ) e2πixξ dξ. Thus in the case of e−πx2we have F ∗(e−πξ

2) = e−πx2

orF ∗ F = I. This is proved later. (Note that F ∗ = F −1 to save a keystroke!).

The Schwartz space of functions comprises those functions that decay rapidly enough so thatthe basic manipulations of Fourier theory work ”nicely”. Indeed, once we are in Schwartz spacewe can do the mathematical equivalent of terrible things to small furry animals without gettingarrested! More specifically, the Schwartz space on R (denoted by S(R)) is the set of all indefi-nitely differentiable functions f so that f and all its derivatves f (l) are rapidly decreasing in thesense that:

supx∈R|x|k | f (l)(x)| < ∞ for all k, l ≥ 0 (1)

Some authors prefer to write (1) in the equivalent form:supx∈R (1 + |x|)k | f (l)(x)| < ∞ for all k, l ≥ 0

That this form is equivalent to (1) can be seen by noting that (1 + |x|)k is simply a polynomialeach of whose terms satisfies (1) and so the finite sum of such terms will also satisfy (1).

This form immediately avoids any issues at x = 0 when performing estimates of the generalform

∫ ∞−∞| f (x)| dx ≤

∫ ∞−∞

ck dx(1+|x|)k

Schwartz did not pull (1) out of the air but unfortunately many expositions of the concept do justthat. Schwartz space S(R) sits between C∞0 (R) and L1(R) such that its functions are invariantunder F ie you stay in Schwartz space under F . If we suppose that f (x) and x f

′

(x) are bothintegrable then we can show that the Fourier transform f (ξ) is differentiable with respect to ξand in fact:d fdξ = F (−2πix f (x) )

Using induction we can then establish that if xn · f (x) is integrable for every integer n > 0,then the Fourier transform is an infinitely differentiable function. Going a bit further, if f is acontinuously differentiable function such that both f (x) and f

′

(x) are integrable and such thatlim|x|→∞ f (x) = 0, then lim|ξ|→∞ ξ · f (ξ) = 0

5

One can think of Schwartz functions either in terms of bounded products as in (1) or as limitproperty:

Limit property - boundedness equivalence

Thus if f ∈ C∞(R) then lim|x|→∞|x|k | f (l)(x)| = 0 for all integers k, l ≥ 0 if and only ifsupx∈R |x|

k | f (l)(x)| < ∞ for all k, l ≥ 0

(1) contains a great deal of information since one can independently fix k and l. The Schwartzspace therefore consists of smooth functions whose derivatives (including the function itself iewhen l = 0) decay at infinity faster than any power. Not only do we have continuity of theSchwarz functions, even better, we have uniform continuity when we choose a closed interval.In (1) when l = 0 we have supx∈R |x|

k | f (x)| < ∞ for all k. This gives the hint of the basicform of f (x) since we know that the exponential ex grows faster than any power of x (and henceits inverse decays correspondingly faster) so it makes sense that a function such as e−x2

, forinstance, inhabits the Schwartz space. It is, of course, infinitely differentiable.

Clearly if p(x) is any polynominal then p(x)e−x2also lives in the Schwartz space because

Schwartz functions fall off at infinity faster than the inverse of any polynomial. Looking at thederivatives of e−x2

we see quickly that they are of the form p(x) e−x2, eg f (2)(x) = (4x2 −2) e−x2

.The following graph shows the boundedness and smoothness of |x|k |(4x2 − 2) e−x2

| for 0 ≤ k ≤10.

It is worth noting here that S(R) ⊆ Lp(R),∀p 1 ≤ p < ∞. This follows since |x|2 | f (x)| → 0 for|x| → ∞. This implies that |x|2p | f (x)|p is bounded, ie ∃M > 0 such that | f (x)|p ≤ M|x|−2p ≤ M

|x|2

for |x| ≥ 1. Thus∫|x|≥1| f (x)|p dx ≤

∫|x|≥1

M|x|2 dx < ∞

Note that e−|x| does not live in the Schwartz space simply because of the lack of differentiabilityat x = 0 notwithstanding the fact that it does fall off rapidly at infinity. Another example of afunction that does not live in Schwartz space is f (x) = 1

(1+|x|2)k since |x|2k

(1+|x|2)k does not decay tozero for any k as |x| → ∞. For instance, when k = 106 and −100000 ≤ x ≤ 100000 the productlooks like this:

6

The function f (x) = e−x2sin(ex2

) is not a creature of Schwartz space because f ′(x) does notdecay to zero as |x| → ∞. This can be seen noting that f ′(x) = 2x cos(ex2

) − 2e−x2x sin(ex2

)so that when x is large the derivative is dominated by the first term whose absolute value isunbounded. The second term goes to zero because xe−x2

→ 0 as x→ ∞.

The Schwartz space concept emerged out of Laurent Schwartz’s rigorous development of dis-tribution theory during the Second World War (how he did this is a story in itself). In the 19th

century the proof techniques were centered around very intricate limit style proofs which in-evitably had to show on a largely case by case that if you took some essentially arbitrary functionf (x) and you multiplied it by e−2πixξ and integrated, you actually got something that converged.Indeed, Fourier’s original (and outrageous) assertion was that any periodic function could berepresented by the series named after him. Thus Fourier believed that you could take his Fouriercoefficients cn = 1

2π

∫ π

−πf (x) e−inx dx and you would get an infinite series

∑n=∞n=−∞ cn einx which

converges to f (x) ie f (x) =∑n=∞

n=−∞ cn einx. This idea met resistance at the time (Lagrange andPoisson were strong opponents of his general approach) and led to many subtle and intricatelimiting arguments as parts of the theory were verified during the 19th century. Dirichlet wasresponsible for many important foundational results. Fourier was wrong in terms of the finedetail but his instincts were right - the Zygmund quotation given at the beginning of this papercaptures the spirit of things - Fourier theory is a delicate matter. Even during the 20th centurythere were doubts about aspects of the theory and it was not until 1966 that Lennart Carlesonshowed that for square integrable functions on [0, 1] the Fourier partial sums converge pointwisealmost everywhere. The proof is difficult to say the least. Carleson actually tried first to disprovethe result and in an interview after he won the 2006 Abel Prize he elaborated on the history ofthe proof [16].

A classic example of how Fourier theory was approached in the 19th century involves the treat-

7

ment of the unit box function defined as follows:

f (x) =

1 if |x| < 12

0 if |x| > 12

(2)

Because f has a discontinuity at x = 12 it was Dirichlet who came up with the idea of replacing

the value of the function at the point of discontinuity by f (x+0)+ f (x−0)2 . Thus in the case of the

unit box function f (± 12 ) = 1

2 . The Fourier transform of the unit box function is the sinc function:

transforms to:

8

Now f is clearly not in Schwartz space since it is discontinuous at x = ± 12 yet it has a bona fide

Fourier transform. The conditions Dirichlet required for the basic theorem were that∫ ∞−∞| f (x)| dx <

∞ and that f and f , are piecewise continuous on every finite interval while at a point of dis-continuity f (x) is replaced by f (x+0)+ f (x−0)

2 . Under these assumptions, at each point where thefunction’s one-sided derivatives exist, the function can be represented by:

f (x) =1π

∫ ∞

0

∫ ∞

−∞

f (ξ) cos[α(ξ − x)] dξ dα (3)

For the details of the intricate limiting arguments that Dirichlet used to establish (3) see Churchill’sbook ( Chapter 6, [17] ). I will not reproduce them here but they were the ”bread and butter” oftraditional Fourier theory courses.

Now if (3) is valid (and it is because f satisfies the hypotheses of the theorem) we should be ableto show that f (±1

2 ) = 12 . Performing the integration we see that:

f (x) =1π

∫ ∞

0

∫ ∞

−∞

f (ξ) cos[α(ξ − x)] dξ dα =1π

∫ ∞

0

∫ 12

− 12

cos[α(ξ − x)] dξ dα

=1π

∫ ∞

0

[sin[α(ξ − x)]α

] 12

− 12

dα =1π

∫ ∞

0

[sin[α( 12 − x)]α

−sin[α(− 1

2 − x)]α

]dα

=1π

∫ ∞

02

sin α2 cosαxα

dα (4)

Hence from (4) we see that:

9

f (12

) =1π

∫ ∞

02

sin α2 cos α

2

αdα =

1π

∫ ∞

0

sinαα

dα =1π

π

2=

12

(5)

Similarly, f (−12 ) = 1

2 . The proof that∫ ∞

0sinαα dα = π

2 can be found in [17, pages 85-86] whichI have set out in expanded form in the Appendix.

In broad terms with the Schwartz space of functions, the focus is on establishing the decay char-acteristics of the functions so that they decay sufficiently fast to damp out the intrinsic oscillatorybehaviour of the exponential term in the integral. Once you define the appropriate rate of decayone can then perform generic limiting arguments with a degree of relative simplicity. It quicklybecomes possible to say with confidence things like ”This will be small because f is rapidly de-creasing” without resort to tedious ε,N style arguments. Indeed, you can compress many stepsin an otherwise detailed proof with such broad references and the chances are you will be ok!Stein and Shakarchi ([2] pages 131-134) briefly set out the foundation for the Schwartz spaceapproach by first considering the space of ”moderately” decreasing functions. These functionsare assumed to be continuous on R and there exists a constant A > 0 such that | f (x)| ≤ A

1+x2 forall x ∈ R.

The problem of convergence has practical dimensions in the context of quantum physics be-cause Feynman’s path integral which arose from Feynman’s investigation of an integral that

looked like this:∫

all space eiε~

m(x−y)2

2ε2 Ψ(y, t) dyA where Ψ(y, t) is the wave function, reflected the basic

characteristics of Fourier-style integrals.

In this paper I prove some basic facts which are essential building blocks for more complex prob-lems. For instance, in the theory of the heat equation you have to evaluate a somewhat dauntinglooking integral of the form

∫ ∞−∞

e(−4π2 ξ2+(1−a)2πiξ)t e2πiξv dξ and some of the techniques exploredbelow are relevant to solving that type of problem (see below for how this particular integral issolved). Later in this paper I deal with the heat equation and its relevance to the solution to theBlack-Scholes equation.

3 Physical considerations

The Gaussian kernel (and the Dirac function for that matter) are not mere mathematical abstrac-tions invented for the delectation of analysts. In fact physics drove the development of the Diracfunction in particular. In advanced physics textbooks there are derivations of the Maxwell equa-tions using microscopic rather than macroscopic principles eg see [section 6.6 of [14] ]. If youfollow the discussion in that book by Jackson you will see that for dimensions large comparedto 10−14m the nuclei can be treated as point systems which give rise to the microscopic Maxwellequations:

10

∇ b = 0, ∇ × e +∂b∂t

= 0, ∇ e =η

ε0, ∇ × b −

1c2

∂e∂t

= µ0j

Here e and b are he microscopic electric and magnetic fields and η and j are the microscopiccharge and current densities. A question arises as to what type of averaging of the microscopicfluctations is appropriate and Jackson says that ”at first glance one might think that averagesover both space and time are necessary. But this is not true. Only a spatial averaging is nec-essary” [[14] , page 249] Briefly the broad reason is that in any region of macroscopic interestthere are just so many nuclei and electrons so that the spatial averaging ”washes” away the timefluctuations of the microscopic fields which are essentially uncorrelated at the relevant distance(10−8m).

The spatial average of F(x, t) with respect to some test function f (x) is defined as:

〈F(x, t)〉 =∫

F(x − x′, t) f (x′) d3x′ where f (x) is real and non-zero in some neighbourhood ofx = 0 and is normalised to 1 over all space. It is reasonable to expect that f (x) is isotropic inspace so that there are no directional biases in the spatial averages. Jackson gives two examplesas follows:

f (x) =

34πR3 , r < R0, r > R

and

f (x) = (πR2)−32 e−

r2

R2

The first example is an average of a spherical volume with radius R but it has a discontinuityat r = R. Jackson notes that this ”leads to a fine-scale jitter on the averaged quantities as asingle molecule or group of molecules moves in or out of the average volume” [ [14], page250]. This particular problem is eliminated by a Gaussian test function ”provided its scale islarge compared to atomic dimensions” [ [14] , p.250]. Luckily all that is needed is that thetest function meets general continuity and smoothness properties that yield a rapidly convergingTaylor series for f (x) at the level of atomic dimensions. Thus the Gaussian plays a fundamentalrole in the calculations presented by Jackson concerning this issue. The article upon whichJackson’s comments are based is that of G Russkaoff in [15]. Jackson gives as an example of thetype of Gaussian test function something like the following [see [14], p.250 Figure 6.1]:

11

-2 -1 1 2

0.2

0.4

0.6

0.8

1.0

The Mathematica code to generate this graph is as follows:

Note that the function in the above graph is infinitely differentiable and is bounded.

4 Some building blocks

Proving the various fundamental building blocks of Fourier transforms on Schwartz functionsnecessarily involves proving that an infinite integral converges. The analysis thus revolvesaround the ”hump” and ”tails” of the integral estimates. Broadly, the tails will be small be-cause the Schwartz function | f (x)| will be rapidly decreasing for large values of x. To get thehump sufficiently small one will usually need to use the fact that the Schwartz functions are uni-formly continuous on closed, bounded intervals. The details will become clearer in the exampleswhich follow.

In what follows we assume that f inhabits the Schwartz space. The Fourier transform of f is:

12

f (ξ) =

∫ ∞

−∞

f (x) e−2πixξ dx (6)

There are other ways to define the Fourier transform. Other possibilities are 1√2π

∫ ∞−∞

f (x)e−iξx dx,∫ ∞−∞

f (x)e−iξx dx and∫ ∞−∞

f (x)e+iξx dx. In quantum physics if ψ(x) is a one dimensional wave

function, its Fourier transform ψ(p) is conventionally defined as 1√2π

∫ ∞−∞

ψ(x) e−ipx dx where

= h2π and h is Planck’s constant. The inverse transform is then 1√

2π

∫ ∞−∞

ψ(p) eipx dp ( see [3],

p.1462] ).

The ”magic” of Fourier theory enables us to get back to f (x) as follows:

f (x) =

∫ ∞

−∞

f (ξ) e2πixξ dξ (7)

That one can do this is due to the Fourier inversion theorem which is proved later.

Let’s start with the Fourier transform of f′

(x) because without this little ”engine” we couldn’t doanything very useful with differential equations. It turns out to be 2πiξ f (ξ). Thus under Fouriertransforms differentiation transforms simply as a product of a multiple of the transformed vari-able and the function’s Fourier transform. So if you have a partial differential equation such as:

∂u∂t

=∂2u∂t2 (8)

where u(x, t) is a function of spatial and temporal dimensions ((8) is in fact the basic heat equa-tion) there are advantages in taking the Fourier transform of both sides with respect to the spatialdimension. Using the rule (twice in the RHS of (8) ) we have not yet proved you get:

∂u(ξ, t)∂t

= −4π2ξ2 u(ξ, t) (9)

Note that the Fourier transform of ∂u(x,t)∂t with respect to x is:

∫ ∞

−∞

∂(x, t)∂t

e−2πixξ dx =∂

∂t

∫ ∞

−∞

u(x, t) e−2πixξ dx =∂

∂tu(ξ, t) (10)

13

That this is the case follows from the rules in relation to differentiating under the integral sign(Leibnitz’s rule) - remember we are in Schwartz space so we can do terrible things to small furryanimals with impunity! More details can be found in the Appendix.

Since (9) is just a garden variety differential equation in t if we fix ξ, you will get u(ξ, t) =

A(ξ) e−4π2ξ2t. To see this use the integrating factor e4π2ξ2t as follows:

∂

∂tu(ξ, t) e4π2ξ2t = e4π2ξ2t

∂u(ξ, t)∂t

+ 4π2ξ2 u(ξ, t)

= 0 (11)

Hence u(ξ, t) e4π2ξ2t = A(ξ) (ie some function independent of t) and so u(ξ, t) = A(ξ) e−4π2ξ2t.Because (8) will involve initial conditions we need to also take the Fourier transform of thoseconditions and we ultimately end with up with:

u(ξ, t) = g(ξ) e−4π2ξ2t (12)

This may not look like it helps but it does because on the RHS of (12) we have a Gaussian and weknow that the Fourier transform of a Gaussian is another Gaussian (generally with a scale factor)so the RHS of (12) is essentially the product of two Fourier transforms and it is a basic result ofFourier theory that the Fourier transform of a convolution of f and g is f (ξ) g(ξ) ie:,

u(ξ, t) = ( f ∗ g)(ξ) =

∫ ∞

−∞

f (ξ − y)g(y) e−2πiyξ dy = f (ξ) g(ξ) (13)

To get back to u(x, t) we use Fourier inversion: F ∗( f ∗ g)(ξ)

= F ∗

F

(u(x, t)

)= u(x, t) . Note

that u(ξ, t) = Fu(x, t)

. In the case of the Black-Scholes equation which has its roots in the heat

equation, using these Fourier transform techniques on the partial differential equation whichlooks like this for 0 < t < T :

∂V∂t

+ rs∂V∂s

+σ2s2

2∂2V∂s2 − rV = 0 (14)

you get a solution that looks like this:

V(s, t) =e−r(T−t)√

2πσ2(T − t)

∫ ∞

0e−(log( s

s∗ )+(r−σ2

2 )(T−t))2

2σ2(T−t) F(s∗)ds∗

s∗(15)

14

5 Proofs of some basic transform properties

1. f′(x)→ 2πiξ f(ξ)

To prove that f′

(x) → 2πiξ f (ξ) just apply the definition of the Fourier transform and use inte-gration by parts. Thus we have:

∫ n

−nf′

(x) e−2πixξ dx =[f (x)e−2πixξ

]n

−n+ 2πiξ

∫ n

−nf (x) e−2πixξ dx (16)

Now as n → ∞, the first term in (16) goes to zero. To see this we have to go back to theproperties of Schwartz functions.

From the definition of Schwartz space we can take l = 0 so that supx∈R |x|k | f (x)| < ∞ for all k ≥ 0.

Because this is a global bound, this means that for any n , 0 we care to choose, ∃s > 0 suchthat |n|k | f (± n)| < s for all k ≥ 0. Thus if we are given any ε > 0 we can choose k large enoughso that s

|n|k < ε, a property which holds for any n we choose. So | f (± n)| < s|n|k < ε for this k. In

other words | f (± n)| → 0 as n→ ∞. Note that at n = 0 we know that f is continuous and henceis bounded on any closed interval so that it does not blow up.

Applying this estimate we see that:

∣∣∣ f (n) e−2π i ξ n − f (−n) e2π i ξ n∣∣∣ ≤ | f (n)| + | f (−n)| < ε + ε (17)

This establishes that[f (x) e−2π i ξ x

]n

−n→ 0 as n → ∞ and so we get the result we wanted. Al-

ternatively, if one uses the limit form of the Schwartz space definition this last estimate comesimmediately since lim|x|→∞| f (x)| → 0 (setting k = 0 in the definition).

Note that the definition of Schwartz space functions allows us to inductively assert that a1|x| | f (x)| <s1, a2|x|2 | f (x)| < s2,...,an|x|n | f (x)| < sn so that |P(x)| | f (x)| < S where P(x) is a polynomial. Thismeans that f decays faster at infinity than the inverse of any polynomial. This fact is used insome later estimates.

2. −2πix f(x)→ ddξ f(ξ)

To prove this we need to establish that f is actually differentiable (which ought to be the casegiven that f lives in the Schwartz space) and actually find the derivative. We start with thisdifference which is the definition of the derivative:

15

f (ξ + h) − f (ξ)h

− ( −2πix f )(ξ) =

∫ ∞

−∞

[ f (x) e−2πix(ξ+h) − f (x) e−2πixξ

h+ 2πix f (x)e−2πixξ

]dx

=

∫ ∞

−∞

f (x) e−2πixξ[e−2πixh − 1

h+ 2πix

]dx = I (18)

It is common in dealing with estimation problems such as that posed by (18) to break the integralup into three parts as follows:(−∞,−N), [−N,N] and (N,∞) and to use relevant properties suchas rapid decrease in the case of Schwartz space functions perhaps in combination with continuity,uniform continuity or differentiabilty as appropriate to ensure that the estimates are sufficientlysmall. In the case of (18) we can use the rapid decrease of f (x) and x f (x) to make the tailsof the integral in (18) sufficiently small. The tails are usually the easiest part of the estimationproblem whereas the middle ”rump” tends to require more subtle estimates. Note here that if fis in S(R) so is x f (x) which follows directly from the definition in (1) where one could simplywrite supx∈R |x|

k−1 |x f (l)(x)| < ∞ for all k ≥ 1, l ≥ 0.

An important property of f is that∫|x|>N |x| | f (x)| dx → 0 as N → ∞. To see this we note

that because f is in Schwartz space it is bounded and rapidly decreasing so ∃B > 0 such that|x|3 | f (x)| < C for some constant C, for all |x| > 1. Thus, |x| | f (x)| < C

|x|2 and so for any ε > 0 wecan find an N such that

∫|x|>N |x| | f (x)| dx <

∫|x|>N

C|x|2 dx = 2C

N < ε.

Our two (the tails are compressed into one) integrals are:

I1 =

∫|x|>N

f (x) e−2πixξ[e−2πixh − 1

h+ 2πix

]dx (19)

I2 =

∫ N

−Nf (x) e−2πixξ

[e−2πixh − 1h

+ 2πix]

dx (20)

Thus:

I = I1 + I2 (21)

so that:

|I| ≤ |I1| + |I2| (22)

Now:

16

|I1| ≤

∫|x|>N

∣∣∣∣ f (x) e−2πixξ[e−2πixh − 1

h+ 2πix

] ∣∣∣∣ dx ≤∫|x|>N| f (x)|

∣∣∣∣ [e−2πixh − 1 + 2πixhh

] ∣∣∣∣ dx

≤ 2π∫|x|>N|x| | f (x)|

|e−2πixh − 1||2πixh|

+ 1

dx (23)

In relation to (23) we know that because of the rapid decrease of f we can make∫|x|≥N | f (x)| |x| dx <

ε because x f (x) is rapdily decreasing. To make∫|x|>N |x| | f (x)|

|e−2πixh−1||2πixh| + 1

dx small enough we

note that for any real θ the following inequality holds: |eiθ − 1| ≤ |θ|. That this is the case can beseen as follows:

|eiθ − 1| = |eiθ2 e

iθ2 − e

iθ2 e−

iθ2 | = |e

iθ2 | |e

iθ2 − e−

iθ2 | = |2 sin

θ

2| ≤ 2

|θ|

2= |θ| (24)

Thus, |e−2πixh−1||2πixh| + 1 ≤ 1 + 1 = 2

Note here that this is a more refined bound on |eiθ − 1| than the obvious one of 2 which wouldlead to an argument about the relative decay of |x| | f (x)| ( 2

2π |x| |h| + 1).

Thus we see that:

|I1| = 2π∫|x|>N |x| | f (x)|

|e−2πixh−1||2πixh| + 1

dx ≤ 4π

∫|x|>N |x| | f (x)| dx < 4πε using the fact that∫

|x|≥N | f (x)| |x| dx < ε.

We now come to |I2| which requires a slightly different line of attack because the interval con-tains the origin and even though f will be bounded at the origin we need to be sure that theintegral is still sufficiently small when N is large. If we fix x and let g(h) = e−2πixh we know thatg′

(h) = −2πix e−2πixh and so g′

(0) = −2πix. Note that g(0) = 1. In fact there is some h0 > 0such that for |h| < h0 we have:

∣∣∣∣e−2πixh − 1h

− (−2πix)∣∣∣∣ =

∣∣∣∣e−2πixh − 1h

+ 2πix∣∣∣∣ < ε

2N(25)

The reason that (25) holds is that e−2πixh−1h + 2πix converges uniformly to zero as h → 0 for all

x ∈ [−N,N]. In essence it is g(h)−g(0)h −g

′

(0) which ought to converge uniformly to zero as h→ 0for any x ∈ [−N,N] because the derivative at issue is continuous on the compact interval [−N,N]and hence uniformly continuous. This can be shown in more detail as follows. Fix x ∈ [−N,N]and take |h| < |h1| < |h0|, and note that by the mean value theorem we can find |h

′

| < |h| suchthat |e

−2πixh−1||2πxh| = |2πxh

′

| and similarly there is a |h′′

| < |h1| such that such that |e−2πixh1−1||2πxh1 |

= |2πxh′′

|.Then

17

∣∣∣∣e−2πixh − 1h

+ 2πix −e−2πixh′ − 1

h′+ 2πix

∣∣∣∣ =∣∣∣∣e−2πixh − 1

h−

e−2πixh′ − 1h′

∣∣∣∣≤

∣∣∣∣e−2πixh − 1h

∣∣∣∣ +∣∣∣∣e−2πixh′ − 1

h′

∣∣∣∣ = |2πx||e−2πixh − 1||2πxh|

+ |2πx||e−2πixh′ − 1||2πxh′|

≤ |2πx| |2πxh′

| + |2πx| |2πxh′′

| ≤ 4π2x2(|h′

| + |h′′|) < 4π2N2 2|h0| = 8π2N2 |h0| < ε (26)

if |h0| <ε

8π2N2 . This establishes the uniform continuity. It is worth noting the subtlety of the

estimate since we know from (24) that |eiθ−1||θ| ≤ 1 which is too big a bound to get the uniform

continuity estimate to work. It is also worth noting that |eiθ−1||θ| = |sinc θ2 | which is uniformly con-

tinuous: http://www.gotohaggstrom.com/Uniform%20continuity%20of%20sinc%20x.pdf

Thus we have:

|I2| =∣∣∣∣ ∫ N

−Nf (x) e−2πixξ

[e−2πixh − 1h

+ 2πix]

dx∣∣∣∣ ≤ ∫ N

−N

∣∣∣∣ f (x)∣∣∣∣ ∣∣∣∣[e−2πixh − 1

h+ 2πix

]∣∣∣∣ dx

≤

∫ N

−NBε

2Ndx = Bε (27)

since ∃B > 0 such that | f (x)| < B on [−N,N]. Putting the two estimates together we have:

|I| ≤ 4πε + Bε < C ε (28)

Hence the result follows. Note that Property 2 is a form of differentiation under the integral signsince it says d

dξ

∫ ∞−∞

f (x) e−2πixξ dx =∫ ∞−∞

(−2πix) f (x) e−2πixξ dx.

3. f(x + h)→ f(ξ) e2πihξ for h ∈ R

To prove this we note that F ( f (x + h) =∫ ∞−∞

f (x + h)e−2πixξ dx and then make the substitutionx′ = x + h so that

∫ ∞−∞

f (x + h)e−2πixξ dx =∫ ∞−∞

f (x′)e−2πi(x′−h)ξ dx′ = f (ξ) e2πihξ.

4. f(x) e−2πixh → f(ξ + h) for h ∈ R

F(f (x)e−2πxh) =

∫ ∞−∞

f (x) e−2πixh e−2πixξ dx =∫ ∞−∞

f (x) e−2πi(x+h)ξ dx = f (ξ + h)

5. f(δx)→ δ−1 f(δ−1ξ) for δ > 0

Consider∣∣∣∣ ∫ N−N f (δx) e−2πixξ dx − δ−1

∫ N−N f (x) e−2πix ξδ dx

∣∣∣∣. Without loss of generality we can as-sume δ > 1. If 0 < δ ≤ 1 the limits of the relevant intervals are reversed.

18

∣∣∣∣ ∫ N

−Nf (δx) e−2πixξ dx −

1δ

∫ N

−Nf (x) e−2πix ξδ dx

∣∣∣∣=

∣∣∣∣1δ

∫ δN

−δNf (x) e−2πix ξδ dx −

1δ

∫ N

−Nf (x) e−2πix ξδ dx

∣∣∣∣=

1δ

∣∣∣∣ ∫ δN

−δNf (x) e−2πix ξδ dx −

∫ N

−Nf (x) e−2πix ξδ dx

∣∣∣∣=

1δ

∣∣∣∣ ∫ −N

−δNf (x) e−2πix ξδ dx +

∫ N

−Nf (x) e−2πix ξδ dx +

∫ δN

Nf (x) e−2πix ξδ dx −

∫ N

−Nf (x) e−2πix ξδ dx

∣∣∣∣=

1δ

∣∣∣∣ ∫ −N

−δNf (x) e−2πix ξδ dx +

∫ δN

Nf (x) e−2πix ξδ dx

∣∣∣∣≤

1δ

∫ −N

−δN| f (x)| dx +

∫ δN

N| f (x)| dx

≤

1δ

∫ −N

−δN

s|x|k

dx +

∫ δN

N

s|x|k

dx

≤2sN(δ − 1)

δNk =2s(δ − 1)δNk−1 → 0 as N → ∞ (29)

Note here that we have used the same logic as applied in the lead up to (17).

What (29) shows is that∫ ∞−∞

f (δx) e−2πixξ dx → 1δ

∫ ∞−∞

f (x) e−2πix ξδ dx ie f (δx)→ δ−1 f (δ−1ξ)

6. If f, g ∈ S(R) then f ∗ g ∈ S(R)

To prove this we need to show that:

supx∈R|x|k |( f ∗ g)(k)(x)| < ∞ ∀k, l ≥ 0 (30)

First start with l = 0 to show that:

supx∈R|x|k |( f ∗ g)(x)| < ∞ ∀k ≥ 0 (31)

To prove (31) the idea is to break up the domain in a way that uses rapid decrease of the functions.First suppose that |x| < 2|y|: Then:

|x|k |g(x − y)| < 2k|y|kB < 2k(1 + |y|)kB ≤ Ak(1 + |y|)k (32)

19

since g ∈ S(R) its is bounded by B, say. Note that Ak = 2kB.

The significance of this estimate is that the RHS of (32) is just a polynomial and we know thata Schwartz space function will decay more rapidly than the inverse of any polynomial and wewill use that fact shortly.

The second step is to assume that |x| ≥ 2|y|. Then:

|x|k |g(x − y)| ≤ |x|kB′

|x − y|k≤ B′

|x|k

|x|k2k

= B′ 2k (33)

where we have used the fact that |x− y| ≥∣∣∣|x| − |y|∣∣∣ ≥ |x|2 because |y| ≤ |x|2 . B′ is a constant.

Thus we have:

|x|k |g(x − y)| ≤ B′ 2k ≤ 2k (1 + |y|)k B′ = Bk (1 + |y|)k (34)

If we choose Ck = maxAk, Bk we can bound |x|k |g(x − y)| by Ck (1 + |y|)k.

So going back to (31) we have:

supx∈R|x|k |( f ∗ g)(x)| = sup

x∈R|x|k

∫ ∞

−∞

f (y) g(x − y) dy∣∣∣ ≤ Ck

∫ ∞

−∞

| f (y)| (1 + |y|)k dy < ∞ ∀k ≥ 0

(35)

That (35) is true follows from the fact that | f (y)| decreases faster than the inverse of any polyno-mial in |y| so the integral is bounded for all k ≥ 0.

Alternatively we can demonstrate the required boundedness in a more detailed way as fol-lows:

∣∣∣∣ ∫ ∞

−∞

f (y) g(x − y) dy∣∣∣∣ ≤ ∣∣∣∣ ∫

|y|≤ |x|2

f (y) g(x − y) dy +

∫|y|> |x|2

f (y) g(x − y) dy∣∣∣∣

≤

∫|y|≤ |x|2

∣∣∣ f (y)∣∣∣ ∣∣∣g(x − y)

∣∣∣ dy +

∫|y|> |x|2

∣∣∣ f (y)∣∣∣ ∣∣∣g(x − y)

∣∣∣ dy

≤

∫|y|≤ |x|2

∣∣∣ f (y)∣∣∣ am

|x − y|m+1 dy +

∫|y|> |x|2

b∣∣∣ f (y)

∣∣∣ dy

≤ am

∫|y|≤ |x|2

∣∣∣ f (y)∣∣∣ 2m+1

|x|m+1 dy + b∫|y|> |x|2

|y|−m |y|m∣∣∣ f (y)

∣∣∣︸ ︷︷ ︸ dy

≤a′

m

|x|m+1

∫|y|≤ |x|2

A dy +b2m

|x|m

∫|y|> |x|2

|y|m∣∣∣ f (y)

∣∣∣︸ ︷︷ ︸ dy

≤cm |x||x|m+1 +

b2m B|x|m

≤Am

|x|m+

Bm

|x|m=

Cm

|x|m(36)

20

This shows that ( f ∗ g)(x) is bounded for all x ∈ R, for all m ≥ 0. Thus f ∗ g ∈ S.

We still have to show that supx∈R|x|l |( f ∗ g)(k)(x)| < ∞ for k > 0. We first establish that:

( ddx

)k( f ∗ g)(x) =

(f ∗

( ddx

)k)(x)∀k = 1, 2 . . . (37)

We know that h(x) =(

ddx

)kg(x) ∈ S(R) if g ∈ S(R) and so we have something of the form

|x|l |( f ∗ g)(x)| where f , g ∈ S(R), hence the above steps demonstrate that this product is rapidlydecreasing for all l ≥ 0.

(37) is established by a simple induction. For k = 1 for have:

ddx

( f ∗ g)(x) =ddx

∫ ∞

−∞

f (y) g(x − y) dy =

∫ ∞

−∞

f (y)ddx

g(x − y) dy = ( f ∗dgdx

)(x) (38)

Now for the induction step:

dk+1

dxk+1 ( f ∗ g)(x) =ddx

(( ddx

)k( f ∗ g)(x)

)=

ddx

∫ ∞

∞

f (y)( ddx

)kg(x − y) dy

=

∫ ∞

∞

f (y)( ddx

)k+1g(x − y) dy = ( f ∗

( ddx

)k+1)(x) (39)

Differentiation under the integral sign is justified by the rapid decrease of the derivatives of g(see the Appendix for more details).

Tying this all together it follows that f ∗ g ∈ S(R).

7. For Schwartz functions, f ∗ g = g ∗ f

( f ∗ g)(x) =

∫ ∞

−∞

f (y) g(x − y) dy =

∫ ∞

−∞

f (x − u) g(u) du = (g ∗ f )(x) (40)

Note that for any Schwartz function F,∫ ∞−∞

F(x) dx =∫ ∞−∞

F(−x) dx because the difference∣∣∣ ∫ N−N F(x) dx −

∫ N−N F(−x) dx

∣∣∣ =∣∣∣ ∫ N−N F(x) dx −

∫ −NN −F(x) dx

∣∣∣ = 0. Moreover F satisfies trans-lation invariance ie ∀h ∈ R,

∫ ∞−∞

F(x− h) dx =∫ ∞−∞

F(x) dx. Thus in the above integration wherethe substitution u = x − y is made, the process is justified since the substitution is a compositionof y→ −y and y→ y − h where h = x.

8. If f, g ∈ S(R) then∫ ∞−∞

f(x) g(x) dx =∫ ∞−∞

f(y) g(y) dy

The proof of this proposition relies upon changing the order of integration for double integrals.Stein and Shakarchi (see [2] page 141) prove this result for the weaker case of moderatelydecreasing functions and it holds true for Schwartz functions.

21

9. Fourier inversion: f ∈ S(R) =⇒ f(x) =∫ ∞−∞

f(ξ) e2πixξ dξ

Proving the inversion formula involves the properties of a Gaussian kernel which I will come to,but the structure of the proof is to first show that:

f (0) =

∫ ∞

−∞

f (ξ) dξ (41)

Having established that result if we let F(y) = f (y + x) we then have:

f (x) = F(0) =

∫ ∞

−∞

F(ξ) dξ =

∫ ∞

−∞

f (ξ) e2πixξ dξ (42)

Noting that f (y + x)→ f (ξ)e2πixξ.

The crux of this proof is the assertion in (42) which necessarily involves a claim of convergenceof the integral for the class of Schwartz functions. If we take a Gaussian kernel Gδ(x) = e−πδx2

then Gδ(ξ) = 1√δe−πξ2δ = Kδ(ξ). Now Kδ(ξ) is a ”good” kernel in the sense that for every

η > 0,∫|x|>η|Kδ(x)| dx→ 0 as δ→ 0. For more on kernels see [12].

Using property 8 stated above we then have:

∫ ∞

−∞

f (x) Kδ(x) dx =

∫ ∞

−∞

f (ξ) Gδ(ξ) dξ (43)

Since Kδ is a good kernel we have that:

∫ ∞

−∞

f (x) Kδ(x) dx→ f (0) as δ→ 0 (44)

One way of seeing this is to note that:

∣∣∣∣ ∫ ∞

−∞

f (x) Kδ(x) dx∣∣∣∣ =

∣∣∣∣ ∫ ∞

−∞

f (x)1√δ

e−πx2δ dx

∣∣∣∣ ≤ ∫ ∞

−∞

| f (y√δ)| e−πy2

dy (45)

Because of the continuity of f , f (y√δ) → f (0) as δ → 0. Hence we get that the RHS of

(45) →∫ ∞−∞

f (0) e−πy2dy = f (0) since

∫ ∞−∞

e−πy2dy = 1. The good kernel is nothing more

than a building block of the Dirac delta ”function” which picks out the value of the function atx = 0.

The continuity property used above can be demonstrated more rigorously. In [ [2] pages 133-134] Stein and Shakarchi show that for a moderately decreasing function f :

∫ ∞

−∞

[ f (x − h) − f (x)] dx→ 0 as h→ 0 (46)

22

This can be generalised to∫ ∞−∞

f (x) Kδ(x) dx → f (0) as δ→ 0 where f is a Schwartz function.To prove this we need to break the domain of the integral into three pieces: a central ”hump”and two symmetrical tails. In the tail pieces the rapid decrease of the Schwartz function is whatgets the estimates small enough. In the hump, one needs uniform continuity in order to get theestimate small enough.

We take N large and fixed and δ > 0. We break the integral up as follows:

∣∣∣∣ ∫ ∞

−∞

[ f (x) − f (0)] Kδ(x) dx∣∣∣∣

=∣∣∣∣ ∫ −N

√δ

−∞

[ f (x) − f (0)] Kδ(x) dx +

∫ N√δ

−N√δ[ f (x) − f (0)] Kδ(x) dx +

∫ ∞

N√δ[ f (x) − f (0)] Kδ(x) dx

∣∣∣∣≤

∫|x|≥N

√δ

∣∣∣∣ f (x) − f (0)∣∣∣∣Kδ(x) dx +

∫ N√δ

−N√δ

∣∣∣∣ f (x) − f (0)∣∣∣∣ Kδ(x) dx (47)

The first integral in (42) represents to the two tails while the other integral is the hump and in itwe make the change of varables y = x√

δ. Then:

∫ N√δ

−N√δ

∣∣∣∣ f (x)− f (0)∣∣∣∣ 1√δ

e−πx2δ dx =

∫ N

−N

∣∣∣∣ f (y√δ)− f (0)

∣∣∣∣ e−πy2dy ≤

∫ N

−N

∣∣∣∣ f (y√δ)− f (0)

∣∣∣∣ dy (48)

Since f is a Schwartz function it is uniformly continuous on any closed interval, say [−N−1,N +

1] so we can choose δ small enough so that supy∈[−N,N]

∣∣∣∣ f (y√δ) − f (0)

∣∣∣∣ ≤ ε4N where ε > 0 is

arbitrary.

Thus∫ N−N

∣∣∣∣ f (y√δ)− f (0)

∣∣∣∣ dy ≤ 2N ε4N = ε

2 . This shows that we can make the hump small enough.

Note that although e−πy2on [−N,N] is bounded by e−πN2

which is small for large N, we havefixed N and it is not enough to say that e−πN2 ∫ N

−N

∣∣∣∣ f (y√δ) − f (0)

∣∣∣∣ dy can be made small becausethe integral could dominate for some small δ. Uniform continuity ensures that if we fix theinterval we can be sure that when the |y

√δ − 0| is small so is

∣∣∣∣ f (y√δ) − f (0)

∣∣∣∣.To estimate the tails we use the boundedness of f no matter what y or δ are:

∫|x|≥N

√δ

∣∣∣∣ f (x)− f (0)∣∣∣∣Kδ(x) dx =

∫|y|≥N

∣∣∣∣ f (y√δ)− f (0)

∣∣∣∣ e−πy2dy ≤ e−πN2

∫|y|≥N

∣∣∣∣ f (y√δ)− f (0)

∣∣∣∣ dy

≤ Be−πN2< ε (49)

10. Convolution with a ”good” kernel Kδ(x) :f ∈ S(R) =⇒ (f ∗Kδ)(x) =

∫ ∞−∞

f(x − t) Kδ(t) dt→ f (x) uniformly as δ→ 0

23

This is a fundamental property and the proof reveals an important interplay between the rapiddecrease of the tails of the kernel and Schwartz functions and uniform continuity in order to getthe estimate for the integral small enough.

A ”good” kernel is represented by a Gaussian of the form:

Kδ(x) =1√δ

e−πx2δ (50)

where δ > 0

We know that∫ ∞−∞

Kδ(x) dx = 1 and we can also show that:

∀η > 0,∫|x|>η|Kδ(x)| dx→ 0 as δ→ 0 (51)

To see why (51) holds we just make te change of variable u = x√δ

so that the intergal becomes∫|u|> η

√δ

e−πu2du which approaches 0 as δ → 0. More precisely, just considering the symmetrical

case of the right tail, if we take the upper limit of the integral as M√δ

where M > 0 is at least as

big as 1√δ

so that as M → ∞, M√δ→ ∞ as δ → 0. By the mean value theorem for the finite

interval [ η√δ, M√

δ], there exists some η

√δ< y < M√

δsuch that

∣∣∣∣ ∫ M√δ

η√δ

e−πu2du

∣∣∣∣ =∣∣∣∣−2πye−πy2 (M−η)

√δ

∣∣∣∣ ≤4πy M√

δe−πy2

→ 0 as y → ∞ due to the fundamental property of the exponential that it growsfaster than any power.

We have to show that∣∣∣∣( f ∗ Kδ)(x) − f (x)

∣∣∣∣ =∣∣∣∣ ∫ ∞−∞ (

f (x − t) − f (x))

Kδ(t) dt∣∣∣∣ → 0 uniformly as

δ→ 0. To do this we break the integral into three pieces: two tails (where |t| > η and one hump(where |t| ≤ η). To estimate the tails we use the rapid decrease of f and the property of Kδ(t)given in (51). Thus for any ε > 0, ∃η > 0 such that | f (x)| < ε

4 for |x| ≥ η

For the hump we use the fact that f is uniformly continuous on any compact (closed, boundedinterval) and the fact that

∫ η

−ηKδ(x) dx < 1. Thus, for any ε > 0, ∃η > 0 such that | f (x)− f (y)| < ε

whenever |x − y| < η

Thus we have:

24

∣∣∣∣( f ∗ Kδ)(x) − f (x)∣∣∣∣ =

∣∣∣∣ ∫ ∞

−∞

(f (x − t) − f (x)

)Kδ(t) dt

∣∣∣∣=

∣∣∣∣ ∫ η

−η

(f (x − t) − f (x)

)Kδ(t) dt +

∫|t|>η

(f (x − t) − f (x)

)Kδ(t) dt

∣∣∣∣≤

∫ η

−η

∣∣∣∣ f (x − t) − f (x)∣∣∣∣ Kδ(t) dt +

∫|t|>η

∣∣∣∣ f (x − t) − f (x)∣∣∣∣ Kδ(t) dt

< e−πη2δ

∫ η

−η

∣∣∣∣ f (x − t) − f (x)∣∣∣∣ dt +

∫|t|>η

(| f (x − t)| + | f (x)|

)Kδ(t) dt

≤ 2ηε e−πη2δ +

∫|t|>η

2ε4

Kδ(t) dt < 2ηε +ε

2< Cε where C is some positive constant (52)

This establishes the uniform convergence as δ→ 0

6 Riemann-Lebesgue Lemma

This is a fundamental result about the decay of the Fourier transform. If f ∈ S(R) then lim|ξ|→∞ f (ξ) =

0. To prove this we need to perform a little fiddle as follows:

f (ξ) =

∫ ∞

−∞

f (x) e−2πixξ dx = −

∫ ∞

−∞

f (x) e−2πiξ(x− 12ξ ) dx

= −

∫ ∞

−∞

f (y +12ξ

) e−2πiyξ dy using the substitution y = x −12ξ

(53)

Therefore:

f (ξ) = −

∫ ∞

−∞

f (x +12ξ

) e−2πixξ dx (54)

(changing the dummy variable from y to x)

and

f (ξ) =

∫ ∞

−∞

f (x) e−2πixξ dx (55)

Adding (54) and (55):

f (ξ) =12

∫ ∞

−∞

[ f (x) − f (x +12ξ

)] e−2πixξ dx (56)

25

To investigate the convergence properties of (56) we perform estimates on the tails (where weuse the rapid decrease of f ) and the hump (where we use uniform continuity). For the tails weknow that ∃B > 0 such that | f (x)| ≤ B

x2 so that we can find an N big enough so that for |ξ| > 1we will have:

∣∣∣∣ ∫|x|>N

[ f (x) − f (x +12ξ

)] e−2πixξ dx∣∣∣∣ ≤ ∫

|x|>N

[| f (x)| + | f (x +

12ξ

)|]

dx

≤

∫|x|>N

[ Bx2 +

B

(x + 12ξ )2

]dx <

ε

2(57)

For the hump where |x| ≤ N the function gξ(x) = f (x) − f (x + 12ξ ) converges uniformly to zero

as |ξ| → ∞ since f is uniformly continuous on [−N,N]. So, given ε > 0, ∃M > 0 such that forall |ξ| > M and all |x| ≤ N, | f (x) − f (x + 1

2ξ )| < ε4N .

Thus putting it all together:

| f (ξ)| ≤∫|x|>N

[| f (x)| + | f (x +

12ξ

)|]

dx +

∫|x|≤N

[ ∣∣∣ f (x) − f (x +12ξ

)∣∣∣ ] dx

≤ε

2+ 2N

ε

4N= ε (58)

This establishes that lim|ξ|→∞ | f (ξ)| = 0

7 If f ∈ S(R) then f is rapidly decreasing

What has to be shown is that:

supξ∈R|ξ|n | f (ξ)| < ∞ for all integers n ≥ 0 (59)

We know from the Riemann-Lebesgue lemma that f vanishes at infinity but we need to show thatas |ξ| → ∞ f vanishes faster than the reciprocal of any polynominal. We also know that:

f′

(x)F−→ 2πiξ f (ξ) (60)

(60) leads inductively to this:

f (n)(x)F−→ (2πiξ)n f (ξ) (61)

26

That (61) holds follows from integration by parts:

∫ ∞

−∞

f (n+1)(x) e−2πixξ dx =[f (n)(x) e−2πixξ

]∞−∞−

∫ ∞

−∞

f (n)(x) (−2πiξ) e−2πixξ dx = (2πiξ)n+1 f (ξ)

(62)

Using the fact that f is a Schwartz function it follows that[f (n)(x) e−2πixξ

]∞−∞

= 0

Similarly, since f (n)(x) ∈ S(R) and the Riemann-Lebesgue lemma we see that:

lim|ξ|→∞

f (n)(ξ) = 0 (63)

ielim|ξ|→∞

(2πiξ)n f (ξ) = 0 (64)

this shows that supξ∈R |ξ|n | f (ξ)| < ∞ which means that f (ξ) is rapidly decreasing.

8 If f ∈ S(R) then f ∈ S(R)

As a first step we note that f ∈ C∞. To establish this we start with the fact that f is a Schwartzfunction so we have that:

∫ ∞

−∞

(1 + |x|)k | f (x)| dx < ∞ (65)

We also have that f is k times differentiable with :

dk

dξk f (ξ) = (−2πix( f (x))k(ξ) (66)

To prove (65) we need to establish the base case k = 1 which follows from Property 2 (the caseof k = 0 is simply the case

∫ ∞−∞| f (x)| dx < ∞ which is a subset of the comments made below)

ie

∫ ∞

−∞

(1 + |x|) | f (x)| dx < ∞ (67)

We know that we can make∫|x|>N | f (x)| dx small since f is rapidly decreasing. For the hump∫

|x|≤N | f (x)| dx we use the fact that f is bounded so the net result is that∫ ∞−∞| f (x)| dx < ∞ .

27

Similarly, for∫|x|>N |x| | f (x)| dx we use rapid decrease properties to ensure the tails are small

enough and for the hump∫|x|≤N |x| | f (x)| dx we use the boundedness of f . To finalise the induction

we note that:

∫ ∞

−∞

(1 + |x|)k+1 | f (x)| dx =

∫ ∞

−∞

(1 + |x|) (1 + |x|)k | f (x)| dx

=

∫ ∞

−∞

(1 + |x|)k | f (x)| dx +

∫ ∞

−∞

|x| (1 + |x|)k | f (x)| dx = A + B < ∞ (68)

A is bounded by the induction hypothesis while B is bounded because we can replicate theearlier arguments. Alternatively, since f is a Schwartz function it decays faster than the inverseof any polynomial so that

∫ ∞−∞

(1 + |x|)k | f (x)| dx is bounded.

9 First proof that e−πx2→ e−πξ

2

To prove that the Fourier transform of the Gaussian f (x) = e−πx2is f (ξ) = e−πξ

2we may as

well use the properties developed above. We start with the definition of the Fourir transformf (ξ) = F(ξ) =

∫ ∞−∞

e−πx2e−2πixξ dx so that F(0) =

∫ ∞−∞

e−πx2dx = 1. Now property 2 established

above ie −2πix f (x)→ ddξ f (ξ) allows us to say that:

F′(ξ) =

∫ ∞

−∞

f (x) (−2πix) e−2πixξ dx = i∫ ∞

−∞

f′

(x) e−2πixξ dx (69)

since f′

(x) = −2πx f (x). We also know from Property 1 ie f′

(x)→ 2πiξ f (ξ) that:

F′

(ξ) = i(2πiξ) f (ξ) = −2πξF(ξ) (70)

So if G(ξ) = F(ξ) eπξ2

then G′

(ξ) = 2πξF(ξ) eπξ2

+ eπξ2

F′

(ξ) and hence G′

(0) = 0 + F′

(0) = 0using (69). This means that G(ξ) is constant for all values of ξ. But since F(0) = 1 we must havethat G is identically equal to 1 which means that F(ξ) = f (ξ) = e−πξ

2

10 Second proof that e−πx2→ e−πξ

2

This is a classically inspired proof using complex variable theory using Cauchy’s Theoremwhich states that if f is holomorphic or analytic in an open set Ω and Γ ⊂ Ω is a closed curvewhose interior is contained in Ω then:

∫Γ

f (z) dz = 0 (71)

28

Now I can hear you saying: ”Where does this get us?”. The answer is that by suitable choiceof f (z) and the contour Γ you can actually get a useful result out of (71). The assumption thatf be holomorphic on the relevant open set is easily satisfied in the case of f (z) = e−πz2

wherez is complex since it is differentiable everywhere in C (If you can’t see why see the Appendix).Thus all we need to do is choose a suitable contour such as the one below so that we can use(71):

Thus we have that:

∫Γ

e−πz2dz = 0 (72)

and we consider the four paths in the contour as follows.

(i) Along AB z = R + iy and dz = i dy where 0 ≤ y ≤ ξ. Hence:

∣∣∣∣ ∫AB

e−πz2dz

∣∣∣∣ =∣∣∣∣ ∫ ξ

0e−π(R2+2iRy−y2) (idy)

∣∣∣∣ ≤ e−πR2∫ ξ

0

∣∣∣∣e−π(2iRy−y2)∣∣∣∣ dy ≤ Cξ e−πR2

(73)

for some constant C noting that∣∣∣∣e−π(2iRy−y2)

∣∣∣∣ =∣∣∣∣e−2πiRy

∣∣∣∣ ∣∣∣∣eπy2∣∣∣∣ ≤ eπξ

2where ξ is fixed. Hence (73)

shows that the integral along AB converges to zeros as R → ∞. Identical logic applies to thepath CD where z = −R + iy and 0 ≤ y ≤ ξ. Thus

∫AB e−πz2

dz =∫

CD e−πz2dz = 0.

Along BC, z = x + iξ where −R ≤ x ≤ R and dz = dx (but note that x starts at R to get the rightorientation for the path so that the interior is on the inside) hence:

∫BC

e−πz2dz =

∫ −R

Re−π(x2+2ixξ−ξ2) dx = −eπξ

2∫ R

−Re−πx2

e−2πixξ dx (74)

Finally, along path DA we have z = x for −R ≤ x ≤ R and dz = dx so that:

29

∫DA

e−πz2dz =

∫ R

−Re−πx2

dx (75)

From (72) we have:∫Γ

e−πz2dz =

∫AB

e−πz2dz +

∫BC

e−πz2dz +

∫CD

e−πz2dz +

∫DA

e−πz2dz = 0 (76)

and using (73)-(75) and letting R→ ∞ (noting that integral in (75) converges to 1) we have:

1 − eπξ2∫ ∞

−∞

e−πx2e−2πixξ dx = 0 =⇒

∫ ∞

−∞

e−πx2e−2πixξ dx = e−πξ

2(77)

11 Applying Fourier transform techniques to solve a variant of theheat equation

The heat equation is such an important equation in physics it is worth understanding some di-mensions to it which are not to my knowledge covered generally in undergraduate engineeringcourses or indeed courses on financial mathematics. What follows is an expansion of some com-ments made by the well known partial differential equation expert Luis Cafarelli of the Univer-sity of Texas, Austin. The heat equation ie ∂u

∂t = ∂2u∂x2 represents a diffusion process. A diffusion

process such as that represented by the heat equation has a tendency to revert to its surroundingaverage. To see how this might be the case we need to look at the most simple situation – ie onedimension, which indicates a relationship between diffusion and the Laplacian (in n dimensionsthe Laplacian is ∆u =

∑ni=1

∂2u∂x2

i).

In one dimension the Laplacian of u is simply the second derivative of u and so we look at thelimit of the second order incremental quotient. Recall that:

u′′

(x) = limh→0

u(x+h)−u(x)h −

(u(x)−u(x−h)

h

)h

= limh→0

u(x + h) − 2u(x) + u(x − h)h2 (78)

In the diagram below the balls represent particles which can jump left and right in proportion totheir number in a pile. The pile at x gains half of the particles coming from adjacent piles andloses its own. This simple rule gives rise to a balance equation of gains (ie 1

2 u(x−h) + 12 u(x + h))

minus losses (ie u(x) )which is proportional to:

12

(u(x + h) + u(x − h) − 2u(x)

)(79)

30

Equation (79) looks suspiciously like (78) - hence the connection with the Laplacian which hasremarkable features: it is rotationally invariant, independent of the system of coordinates andrepresents a diffusion. As we go up in dimensions we consider the Laplacian as a limit gain-lossof density u at x. We take the average over a unit sphere S of the radial second derivatives inevery direction and one of the fundamental results of harmonic analysis is that:

∆(u) =

?S

urr dA(s) (80)

Recall that for a function u defined in a ball B(x, r) of radius r about x in Rn, with boundary∂B(x, r) and α(n) is the volume of a unit ball in Rn and nα(n) is the surface area of the unit ballin Rn, the average of u on B(x, r) is:?

B(x,r)u(y) dy =

1α(n)rn

∫B(x,r)

u(y) dy (81)

In 2 dimensions a function u is harmonic at P (ie it satisfies Laplace’s equation ∆ = 0) if andonly if:

u(P) =1

2πr

∫∂B(P,r)

u ds =1πr2

∫B(P,r)

u dx dy (82)

To prove (82) we take P = (x0, y0) and we suppose that u(x0, y0) = 12πr

∫∂B(P,r) u(x, y) ds then:

u(x0, y0) =1

2πr

∫∂B(P,r)

u(x, y) ds =1

2πr

∫ 2π

0u(x0 + r cos θ, y0 + r sin θ) rdθ

=1

2π

∫ 2π

0u(x0 + r cos θ, y0 + r sin θ) dθ (83)

The LHS of (83) is simply a constant so if we differentiate with respect to r under the integralsign (and use the chain rule) we get:

31

0 =1

2π

∫ 2π

0(∂u∂x

cos θ +∂u∂y

sin θ) dθ =1

2πr

∫ 2π

0(∂u∂x

cos θ +∂u∂y

sin θ) rdθ

=1

2πr

∫∂B(P,r)

∇u ν ds =1

2πr

∫B(P,r)

div (∇u) dy dx =1

2πr

∫B(P,r)

∆u dy dx (84)

The divergence theorem justifies the last step in (84). Hence based on our assumption u(x0, y0) =1

2πr

∫∂B(P,r) u(x, y) ds we have shown that 0 =

∫B(P,r) ∆u dy dx for all r > 0. If all this holds for

every P in some open subset Ω in R2 then we must have that ∆u = 0 for each such P. Thus u isharmonic.

What this averaging suggests is that the heat equation ∂u∂t = ∆(u) reflects the fact that the

density u at the point x makes an infinitesimal comparison within its neighbourhood andtries to revert to the surrounding average.

Problem 1 at pages 169-170 of [2] reads as follows:

The equation:

x2 ∂2u∂x2 + ax

∂u∂x

=∂u∂t

(85)

with u(x, 0) = f (x) for 0 < x < ∞ and t > 0 is a variant of the heat equation. This equationcan be solved by making the change of variable x = e−y so that −∞ < y < ∞. If we setU(y, t) = u(e−y, t) and F(y) = f (e−y) then (85) becomes:

∂2U∂y2 + (1 − a)

∂U∂y

=∂U∂t

(86)

with U(y, 0) = F(y). One then has to show that the solution to the original problem is:

u(x, t) =1√

4πt

∫ ∞

0e−(ln( v

x )+(1−a)t)2

4t f (v)dvv

(87)

We start with:

∂U∂y

=∂u∂x

∂x∂y

+∂u∂t

∂t∂y

=∂u∂x× (−e−y) = −x

∂u∂x

(88)

∂2U∂y2 =

∂

∂y(∂U∂y

)=

∂

∂x(− x

∂u∂x

) ∂x∂y

+ −x∂u∂x

∂t∂y

= −x(− x

∂2u∂x2 +

∂u∂x× −1

)= x2 ∂

2u∂x2 + x

∂u∂x

(89)

32

(1 − a)∂U∂y

= −x(1 − a)∂u∂x

(90)

∂U∂t

=∂u∂y

∂y∂t

+∂u∂t∂t∂t

=∂u∂t

(91)

Using (89)-(91) we have:

∂2U∂y2 + (1 − a)

∂U∂y

= x2 ∂2u∂x2 + x

∂u∂x− x(1 − a)

∂u∂x

= x2 ∂2u∂x2 + ax

∂u∂x

(92)

and since ∂U∂t = ∂u

∂t the transformation from (85) to (86) is established.

We now take Fourier transforms of (86) with respect to y as well as the related initial conditions.The Fourier transform of (34) with respect to y is:

−4π2ξ2U(ξ, t) + (1 − a)2πiξ U(ξ, t) =∂U(ξ, t)∂t

(93)

Note that the Fourier transform of ∂U(y,t)∂t with respect to y is:

∫ ∞

−∞

∂U(y, t)∂t

e−2πiyξ dy =∂

∂t

∫ ∞

−∞

U(y, t) e−2πiyξ dy =∂

∂tU(ξ, t) (94)

See the Appendix for more details about differentiating under the integral sign.

The Fourier transform of the initial condition U(y, 0) = F(y) is U(ξ, 0) = F(ξ).

To solve (93) we use the integrating factor e−[−4π2ξ2+(1−a)2πiξ]t = e−Φ(ξ)t where Φ(ξ) = −4π2ξ2 +

(1 − a)2πiξ. Thus (93) becomes:

Φ(ξ) U(ξ, t) −∂U(ξ, t)∂t

= 0 (95)

and multiplying by the integrating factor we have:

e−Φ(ξ)t[Φ(ξ) U(ξ, t) −

∂U(ξ, t)∂t

]= 0 (96)

But (96) is equivalent to:

∂

∂t

e−Φ(ξ)t U(ξ, t)

= 0 (97)

33

This means that:

e−Φ(ξ)t U(ξ, t) = A(ξ) (98)

where A(ξ) is some function independent of t.

But U(ξ, 0) = F(ξ) therefore A(ξ) = U(ξ, 0) = F(ξ)

So:

U(ξ, t) = F(ξ) eΦ(ξ)t (99)

At this point, if we know what function transforms to eΦ(ξ)t then (99) says that U(ξ, t) is the prod-uct of two transforms and we can recover the original U(y, t) and hence u(x, t) by convolution ieif h(ξ) = f (ξ) g(ξ) then h(x) =

∫ ∞−∞

f (x − t) g(t) dt

We know that the function which transforms to eΦ(ξ)t must be a Gaussian with appropriate scalingand the detailed derivation is given below.

We let G(ξ, t) = eΦ(ξ)t =∫ ∞−∞

G(v, t) e−2πivξ dv

Therefore, by Fourier Inversion:

G(v, t) =

∫ ∞

−∞

eΦ(ξ)t e2πivξ dξ =

∫ ∞

−∞

e[−4π2ξ2+(1−a)2πiξ]t e2πivξ dξ (100)

To evaluate (100) we need to complete the square in the exponential factor. Thus we have:

∫ ∞

−∞

e[−4π2ξ2+(1−a)2πiξ]t e2πivξ dξ =

∫ ∞

−∞

e−4π2tξ2+(1−a)2πiξt+2πivξ dξ =

∫ ∞

−∞

e−4π2tξ2− 2πi

4π2t[(1−a)t+v]ξ

dξ

=

∫ ∞

−∞

e−4π2t

[ξ− i

4πt [(1−a)t+v]]2

+[(1−a)t+v]2

(4πt)2

dξ = e

−[(1−a)t+v]24t

∫ ∞

−∞

e−4π2tξ− i

4πt [(1−a)t+v]2

dξ

= e−[(1−a)t+v]2

4t

∫ ∞+ i4πt [(1−a)t+v]

−∞− i4πt [(1−a)t+v]

e−4π2tz2dz

(101)

To evaluate the integral in (101) one can use the contour given in section 10 in the context ofshowing that the Gaussian is its own Fourier transform and when you reproduce the steps youwill find that:

e−[(1−a)t+v]2

4t

∫ ∞+ i4πt [(1−a)t+v]

−∞− i4πt [(1−a)t+v]

e−4π2tz2dz = e

−[(1−a)t+v]24t

1√

4πt(102)

34

Another way of ”doing” the integral in (101) (apart from using Mathematica or Matlab!) is toview it as basically a real integral with real limits since the imaginary part in the limits contributea negligible vertical displacement (note that t and v are treated as fixed) so it is effectively thisintegral (treating x as a real variable):

∫ ∞−∞

e−4π2tx2dx = 1√

4πtsince

∫ ∞−∞

e−πx2dx = 1. Just think

of the crude image of the infinite real integral being rotated a small amount about the origin, Ofcourse the contour integration approach set out in detail in section 5 is the proper way to do theintegral and all I am suggesting here is a way of remembering what the result is by way of auseful visualisation.

Going back to (99) we now know what functions transform to F(ξ) and eΦ(ξ)t, namely, F(y)(recall that the initial condition was U(y, 0) = F(y) and we took the Fourier transform) and

e−[(1−a)t+v]2

4t 1√4πt

. To get back to U(y, t) we take the inverse Fourier transform of (99) which meansthat we will get the convolution the two functions just mentioned. More specifically we willget:

U(y, t) =

∫ ∞

−∞

G(y − v) F(v) dv =

∫ ∞

−∞

e−[(1−a)t+y−v]2

4t1√

4πtF(v) dv (103)

Now recalling that x = e−y so that ln x = −y we also make the substiution v∗ = e−v so thatln v∗ = −v and dv∗ = −e−vdv = −v∗dv and F(v) = f (e−v) = f (v∗) (go back to the conditionsrelating to (85) and the integral in (103) becomes:

∫ ∞

−∞

e−[(1−a)t+y−v]2

4t1√

4πtF(v) dv =

∫ 0

∞

e−[(1−a)t−ln x+ln v∗]2

4t1√

4πt

f (v∗)−v∗

dv∗

=1√

4πt

∫ ∞

0e−[(1−a)t+ln

(v∗x

)]2

4tf (v∗)v∗

dv∗ = u(x, t) (104)

Thus we get to the advertised solution given in (87) (the variable v∗ replaces v).

12 Using Fourier transform techniques to solve the Black-Scholesequation

The approach to the solution of the problem given in section 11 provides the foundation forsolving the Black-Scholes partial differential equation from finance. The original derivation (see[10]) did not explicitly involve Fourier transform techniques. Indeed, the authors merely referredto a well-known undergraduate Fourier series textbook for the solution to what is essentially aheat equation.

Problem 1 at page 170 of [2] sets up the Black-Scholes differential equation so it can be attackedby the Fourier transfrom techniques already covered. The partial differential equation is:

35

∂V∂t

+ rs∂V∂s

+σ2s2

2∂2V∂s2 − rV = 0, 0 < t < T (105)

This equation is subject to a ’final’ boundary condition which is:

V(s,T ) = F(s) (106)

It will be seen that this final condition is essentially an initial condition once we have done anappropriate transformation. With some experience in playing with differential equations such as(105) a reasonable substitution would look like this:

V(s, t) = eax+bτ U(x, τ) (107)

where x = ln s and τ = σ2

2 (T − t), a = 12 −

rσ2 and b = −( 1

2 + rσ2 )2. Although these parameters

are given in the problem in what follows they are derived so you can see where they come from.With these substitutions (105) can be reduced to a simple one dimensional heat equation withan initial condition U(x, 0) = e−ax F(ex) which can then be attacked using the Fourier transformtechniques explored above. The problem asks you to prove that:

V(s, t) =e−r(T−t)√

2πσ2(T − t)

∫ ∞

0e−(log( s

s∗ )+(r−σ2

2 )(T−t))2

2σ2(T−t) F(s∗)ds∗

s∗(108)

In relation to (108) which is problem 2 in [2], page 170 there is a typo - the 1s∗ factor is missing.

This correction will become clear in the derivation.

Under the substitution x = ln s and τ = σ2

2 (T − t) we can get expressions for the relevant partialderivatives as follows:

∂V∂x

=∂V∂s

∂s∂x

+∂V∂τ

∂τ

∂x(109)

But ∂τ∂x = 0 and ∂s

∂x = s so (109) becomes:

∂V∂x

= s∂V∂s

= Ω (110)

The second partial derivative is:

∂2V∂x2 =

∂Ω

∂x=∂Ω

∂s∂s∂x

+∂Ω

∂τ

∂τ

∂x=∂Ω

∂s∂s∂x

= s∂

∂s

(s∂V∂s

)= s2 ∂

2V∂s2 + s

∂V∂s

= s2 ∂2V∂s2 +

∂V∂x

(111)

36

Thus we have:

s2 ∂2V∂s2 =

∂2V∂x2 −

∂V∂x

(112)

We now do the same for the time derivative τ:

∂V∂τ

=∂V∂s

∂s∂τ

+∂V∂t

∂t∂τ

=∂V∂t

∂t∂τ

=−2σ2

∂V∂t

(113)

since ∂s∂τ = 0 and ∂t

∂τ = ∂∂τ T −

2σ2 τ = −2

σ2

Hence, using ((110)-(113), (105) becomes:

−σ2

2∂V∂τ

+ r∂V∂x

+σ2

2

(∂2V∂x2 −

∂V∂x

)− rV = 0

that isσ2

2∂V∂τ

=σ2

2∂2V∂x2 +

(r −

σ2

2) ∂V∂x− rV (114)

Now with the substiution V = eax+bτ U(x, τ) the relevant derivatives become:

∂V∂τ

= beax+bτU + eax+bτ ∂U∂τ

= eax+bτ (bU +∂U∂τ

)(115)

∂V∂x

= eax+bτ ∂U∂x

+ aeax+bτ U = eax+bτ (aU +∂U∂x

)(116)

∂2V∂x2 = eax+bτ

(∂2U∂x2 + a

∂U∂x

)+ aeax+bτ

(aU +

∂U∂x

)= eax+bτ

(∂2U∂x2 + 2a

∂U∂x

+ a2U)

(117)

Substituting (115)-(117) into (114) we get:

σ2

2eax+bτ (bU+

∂U∂τ

)=σ2

2eax+bτ

(∂2U∂x2 +2a

∂U∂x

+a2U)+(r−σ2

2)

eax+bτ (aU+∂U∂x

)−reax+bτU

(118)

On dividing by the exponential factor we are left with:

σ2

2∂U∂τ

=σ2

2∂2U∂x2 +

(aσ2 + r −

σ2

2)∂U∂x

+[a2σ2

2+ a(r −

σ2

2) − r −

bσ2

2]U (119)

37

To make (119) into a one dimensional heat equation we need:

aσ2 + r −σ2

2= 0 (120)

and

a2σ2

2+ a(r −

σ2

2) − r −

bσ2

2= 0 (121)

(120) shows that a = 12 −

rσ2 as given in the statement of the problem. Substituting this value for

a into (121) gives:

σ2

2(14−

rσ2 +

r2

σ4

)+

(12−

rσ2

) (r −

σ2

2)− r −

bσ2

2= 0

σ2

8−

r2

+r

2σ2 +r2−σ2

4−

r2

σ2 +r2− r =

bσ2

2bσ2

2= −

(σ2

8+

r2

+r2

2σ2

)b = −

(14

+rσ2 +

r2

σ4

)= −

(12

+rσ2

)2 (122)

Accordingly, (119) becomes (after dividing through by σ2

2 ):

∂U∂τ

=∂2U∂x2 (123)

which is the familiar one dimensional heat equation. The initial condition for this equation isfound by putting τ = 0 and hence T = t in (107) giving:

V(s,T ) = eax U(x, 0) (124)

But V(s,T ) = F(s) = F(ex) so the initial condition becomes:

U(x, 0) = e−ax F(ex) (125)

Taking the Fourier transform of (123) with respect to x we get:

∂U(ξ, τ)∂τ

= −4π2 ξ2 U(ξ, τ) (126)

38

Taking the Fourier transform of the initial condition (125) with respect to x we get:

U(ξ, 0) = Fe−ax F(ex)

= C(ξ) (127)

As before the solution to (126) is:

U(ξ, τ) = C(ξ) e−4π2 ξ2 τ = U(ξ, 0) e−4π2 ξ2 τ (128)

Equation (128) is thus the product of two Fourier transforms and so we can get back to U(x, τ)and hence V(s, t) by convolution of the functions which give rise to those transforms. TheGaussian transform in (128) is easy since we know that a Gaussian transforms to a Gaussianso we only have to get the scaling right. But we know that if f (x) = e−πx2

then f (ξ) = e−πξ2

and f (δ x) → 1δ f

( ξδ

)so that by letting δ = 1√

4π τwe can see that e−4π2 ξ2 τ is the transform of

1√4πτ

e−x24τ . The other function is even easier to identify since from (125) the function which we

want is simply e−ax F(ex)

Thus by convolution we have:

U(x, τ) =1√

4πτ

∫ ∞

−∞

e−ay F(ey) e−(x−y)2

4τ dy (129)

Now make the folowing substitutions:x = ln s for 0 < s < ∞y = ln s∗ for 0 < s∗ < ∞ so that dy = ds∗

s∗

τ = σ2

2 (T − t) as beforeV(s, t) = eax+bτ U(x, τ) as before

Using (107) we therefore have:

39

V(s, t) = eax+bτ U(x, τ) =ea ln s+bσ2

2 (T−t)√2πσ2(T − t)

∫ ∞

0e−a ln s∗ F(s∗) e

−[ln( ss∗ )]2

2σ2(T−t)ds∗

s∗

=ebσ2

2 (T−t)√2πσ2(T − t)

∫ ∞

0F(s∗) ea ln( s

s∗ ) e−[ln( s

s∗ )]2

2σ2(T−t)ds∗

s∗

=ebσ2

2 (T−t)√2πσ2(T − t)

∫ ∞

0e

−12σ2(T−t)

[[ln( s

s∗ )]2–2aσ2(T−t) ln( ss∗ )

]F(s∗)

ds∗

s∗

=ebσ2

2 (T−t)√2πσ2(T − t)

∫ ∞

0e

a2σ2(T−t)2 e

−12σ2(T−t)

[ln( s

s∗ )–( 12−

rσ2 )σ2(T−t)

]2

F(s∗)ds∗

s∗

=eσ22 (T−t)(a2+b)√2πσ2(T − t)

∫ ∞

0e

−12σ2(T−t)

[ln( s

s∗ )+(r−σ2

2 )(T−t)]2

F(s∗)ds∗

s∗

=eσ2(T−t)

2 × −2rσ2√

2πσ2(T − t)

∫ ∞

0e

−12σ2(T−t)

[ln( s

s∗ )+(r−σ2

2 )(T−t)]2

F(s∗)ds∗

s∗

=e−r(T−t)√

2πσ2(T − t)

∫ ∞

0e

−12σ2(T−t)

[ln( s

s∗ )+(r−σ2

2 )(T−t)]2

F(s∗)ds∗

s∗

(130)

recalling that a2 + b = ( 12 −

rσ2 )2 − ( 1

2 + rσ2 )2 = −2r

σ2

It is not immediately obvious that V(s, t) = e−r(T−t)√

2πσ2(T−t)

∫ ∞0 e

−12σ2(T−t)

[ln( s

s∗ )+(r−σ2

2 )(T−t)]2

F(s∗) ds∗s∗ is

in the form given in standard options pricing textbooks such as [12, page 180]. For instance, thegeneral form of the Black-Scholes formula for the value of a European call option C struck at Kon stock worth S and expiring at time T (assuming no dividends) is:

C = N(d1)S − e−r(T−t) N(d2)K (131)

where:

N(z) =1√

2π

∫ z

−∞

e−x22 dx (132)

d1 =ln( S

K ) + (r + σ2

2 )(T − t)

σ√

T − t(133)

d2 =ln( S

K ) + (r − σ2

2 )(T − t)

σ√

T − t(134)

40

To get from (130) to (131) we have to get an expression for F(s∗) and do a change of variablesmaking sure that the limits of integration are appropriately adjusted.

We make the substitution:

u =ln( s

s∗ ) + (r − σ2

2 )(T − t)

σ√

T − tso that −σ

√T − t du =

ds∗

s∗(135)

The limits of integration s∗ = 0→ u = ∞ and s∗ = ∞ → u = −∞

Note that:

s∗ = s e−[uσ√

T−t− (r−σ2

2 )(T−t)] (136)

From the statement of the problem we know that V(s∗,T ) = F(s∗) and for a European call optionF(s∗) = maxs∗ − 1, 0. Note here that in terms of the general situation the relevant condition ismaxs∗ − K, 0 but in the development of this problem K = 1 implicitly. This is so because theoriginal change of variables was x = ln s but in the more general context s = Kex.

At this stage we have this:

V(s, t) =e−r(T−t)√

2πσ2(T − t)

∫ −∞

∞

e−u2

2 F(s e−[uσ√

T−t− (r−σ2

2 )(T−t)])× −σ

√T − t du (137)

Now F(s e−[uσ√

T−t− (r−σ2

2 )(T−t)]) = maxs e−[uσ√

T−t− (r−σ2

2 )(T−t)] − 1, 0 so that ifF(s e−[uσ

√T−t− (r−σ

22 )(T−t)]) = maxs e−[uσ

√T−t− (r−σ

22 )(T−t)] − 1 > 0 we can write the integral as

follows:

V(s, t) =e−r(T−t)√

2π

∫ ln s+(r−σ2

2 )(T−t)

σ√

T−t

−∞

e−u2

2s e−[uσ

√T−t− (r−σ

22 )(T−t)] − 1

du (138)

For maxs e−[uσ√

T−t− (r−σ2

2 )(T−t)]−1 > 0 to hold we need s e−[uσ√

T−t−(r−σ2

2 )(T−t)] > 1 or e−[uσ√

T−t− (r−σ2

2 )(T−t)] >1s . Taking logs we have that:

−[uσ√

T − t − (r − σ2

2 )(T − t)] > ln( 1s ) = − ln s =⇒ uσ

√T − t − (r − σ2

2 )(T − t) < ln s and so

u < ln s+(r−σ2

2 )(T−t)

σ√

T−t. This gives the upper limit for u.

We can now write (138) as two integrals ie V(s, t) = I − J where:

I =e−r(T−t)√

2π

∫ ln s+(r−σ2

2 )(T−t)

σ√

T−t

−∞

e−u2

2 s e−[uσ√

T−t−(r−σ2

2 )(T−t)] du (139)

41

and

J =e−r(T−t)√

2π

∫ ln s+(r−σ2

2 )(T−t)

σ√

T−t

−∞

e−u2

2 du (140)

Using the normal distribution function N(.) we can write J = e−r(T−t) N(d2). Note that this isconsistent with (131) with K = 1 as has been assumed in this derivation.

To simplify I we have to complete the square and then do another change of variable. Theexponential term has exponent:

−u2

2− [uσ

√T − t − (r −

σ2

2)(T − t)] =

−[u2 + 2uσ√

T − t − 2(r − σ2

2 )(T − t)]2

=−[(u + σ

√T − t)2 − σ2(T − t) − 2(r − σ2

2 )(T − t)]2

=−(u + σ

√T − t)2

2+ r(T − t) (141)

Thus:

I =e−r(T−t)√

2π

∫ ln s+(r−σ2

2 )(T−t)

σ√

T−t

−∞

s e−(u+σ

√T−t)2

2 er(T−t) du

=1√

2π

∫ ln s+(r−σ2

2 )(T−t)

σ√

T−t

−∞

s e−(u+σ

√T−t)2

2 du (142)

We now make the change of variable:

z = u + σ√

T − t so that dz = du (143)

The upper limit in I becomes:

z =ln s + (r − σ2

2 )(T − t)

σ√

T − t+ σ√

T − t =ln s + (r − σ2

2 )(T − t) + σ2(T − t)

σ√

T − t

=ln s + (r + σ2

2 )(T − t)

σ√

T − t(144)

So from(142) I becomes:

42

I =s√

2π

∫ ln s+(r+σ2

2 )(T−t)

σ√

T−t

−∞

e−z2

2 dz = s N(d1) (145)

Hence:

V(s, t) = sN(d1) − e−r(T−t) N(d2) (146)

which is the same as (131) with K = 1. There are, of course, other ways to get to the same resultbut they also involve many similar error prone steps.

13 References

[1] R Franklin and R Gosling, ”The Molecular Structure of Sodium Thymonucleat”, Nature,April 25, 1953, pages 740-741

[2] Werner Heisenberg, ”The Physical Principles of the Quantum Theory”, Dover 1949.

[3] Elias M Stein & Rami Shakarchi, Fourier Analysis: An Introduction, Princeton UniversityPress, 2003

[4] Claude Cohen-Tannoudji, Bernard Diu & Franck Laloe, Quantum Mechanics, Volume 2 ,John Wiley & Sons, 1977

[5] Elias M Stein & Rami Shakarchi, Complex Analysis, Princeton University Press, 2003, pages14-18.

[6] T. W. Korner, ”Fourier Analysis”, Cambridge University Press, 1988.

[7] Frank Jones, ”Lebesgue Integration on Euclidean Space”, Bartlett and Jones Publishers, 2001,page 154-55

[8] Murray M Spiegel, ”Schaum’s Outline Series: Theory and Problems of Advanced Calculus”,McGraw Hill, 1963, p.170

[9] Erik Tavila, Necessary and Sufficient Conditions for Differentiating under the Integral Sign,American Mathematcal Monthly, Vol 108, No 6, June-July 2001, pages 544-548, http://www.jstor.org/stable/2695709

[10] Harley Flanders Differentiating under the Integral Sign, American Mathematical Monthly,Vol 80 No 6, June-July 1973, pages 615-627, http://www.jstor.org/stable/2319163

[11] Fischer Black and Myron Scholes,”The Pricing of Options and Corporate Liabilities”, TheJournal of Political Economy, Vol. 81, No. 3 (May - Jun., 1973), pp. 637-654: http://www.cs.princeton.edu/courses/archive/fall08/cos323/papers/black_scholes73.pdf

43

[12] Neil A Chriss, Black-Scholes and Beyond - Option Pricing Models,McGraw-Hill, 1997

[13] Peter Haggstrom, ”The good, the bad and the ugly of kernels:Why the Dirichlet kernel is not a good kernel ”, http://www.gotohaggstrom.com/The%20good,%20the%20bad,%20and%20the%20ugly%20of%20kernels.pdf

[14] John David Jackson, Classical Electrodynamics , Third Edition, John Wiley, 1999

[15] G Russakoff, “ A Derivation of the Macroscopic Maxwell Equations”, Am. J. Physics. 38,1188 (1970). This article can be viewed via this link: http://atsol.fis.ucv.cl/dariop/sites/atsol.fis.ucv.cl.dariop/files/A_Derivation_of_the_Macroscopic_Maxwell_

Equations_-_G._Russakoff.pdf

[16] Martin Raussen and Christian Skau, “Interview with Abel Prize Recipient Lennart Carleson,”Notices of the American Mathematical Society, February 2007, pages 223-229 The interviewcan be viewed online via this link: http://www.ams.org/notices/200702/comm-carleson.pdf

[17] Ruel V Churchill, “Fourier Series and Boundary Value Problems”, Second Edition,McGrawHill, 1963

[18] T J I’A Bromwich, “An Introduction to the Theory of Infinite Series”, Merchant Books1908 (digitized by Watchmaker Publishing 2008)

[19] W T Gowers, “A New Proof of Szemeredi’s Theorem”, https://www.dpmms.cam.ac.uk/˜wtg10/papers.html

[20] Elias M Stein & Rami Shakarchi, ”Functional Analysis: Introduction to Further Topics in Analysis”,Princeton University Press, 2011

[21] Jyrki Kauppinen and Jar Partanen, ”Fourier Transforms and Spectroscopy”, Wiley,2001,page 18

[22] A H Zemanian, ”Distribution Theory and Transform Analysis: An Introduction to Generalized Functions with Applications”,Dover Publications, 1965

14 APPENDIX

14.1 Proof that ez2is differentiable for all z ∈ C

Using power series techniques one can show that ez is differentiable for all z ∈ C (see [4 pages14-18]). Using the chain rule which states that if d : Ω → U and g : U → C are holomorphicthen (g f )

′

(z) = g′

( f (z)) f′

(z) for all z ∈ Ω we can differentiate ez2to get 2z ez2

since z2 is clearly

44

differentiable (holomorphic) for all z ∈ C because the limit (z+h)2−z2

h = 2z + h exists everywhereas h→ 0.

Alternatively, one can use the Cauchy-Riemann conditions to prove the differentiability on C.Thus if x = x + iy then ez2

= ex2−y2cos 2xy + i ex2−y2

sin 2xy = u(x, y) + iv(x, y) and a straigh-forward calculation shows that ∂u

∂x = ∂v∂y and ∂v

∂x = −∂u∂y where u(x, y) and v(x, y) are continuously

differentiable on C. Thus the derivative exists on C ie the function is holomorphic on C.

14.2 Differentiating under the integral sign

If we assume that U(y, t) lives in Schwartz space it is an easy matter to show that we can differ-entiate under the integral sign. A basic theorem underpinning differentiation under the integralsign in a Fourier theory context is given in [6, pp.268-9] and runs as follows.

14.2.1 Main Result

Let g : R × R → C be a continuous function such that ∂g(x,t)∂t exists and is continuous. Suppose∫ ∞

−∞|g(x, t)| dx and

∫ ∞−∞|∂g(x,t)∂t | dx and exist for each t and that

∫|x|>R

∣∣∣∣∂g(x,t)∂t

∣∣∣∣ dx → 0 as R → ∞

uniformly in t on every interval [a, b] . Then∫ ∞−∞

g(x, t) dx is differentiable with:

ddt

∫ ∞−∞

g(x, t) dx =∫ ∞−∞

∂g(x,t)∂t dx

If the function inhabits Schwartz space the continuity, uniform continuity and integral existenceassumptions will be satisfied because of the rapid decrease characteristics of the function.

14.2.2 Using Lebesgue integration theory

To prove the Main Result requires a number of steps that involve continuity, uniform continuity,pointwise convergence and the Fundamental Theorem of Calculus. The process is quite intricateand builds on several results that are usually covered in first and second year Analysis courses.There are other approaches based on Lebesgue measure theory which can yield swifter proofsrelying basically on the Lebesgue Dominated Convergence Theorem . For instance, in [6] thetheorem is stated this way: Assume that for each x ∈ J, f (x, t) is an integrable function of xon X. Assume that for µ a.e. (almost everywhere) x ∈ X, f (x, t) is differentiable at all t ∈ J.Moreover, assume that there exists h ∈ L1(X) such that the derivative satisfies for µ a.e. x ∈ Xand all t ∈ J,

∣∣∣∣d f (x,t)dt

∣∣∣∣ ≤ h(x).

Then F is also differentiable on J and dF(t)dt =

∫X

d f (x,t)dt dµ(x).

The proof runs like this. Let t ∈ J be fixed and for small |δ| consider:F(t+δ)−F(t)

δ =∫

Xf (x,t+δ)− f (x,t)

δ dµ(x)

45

The integrand is continuous at δ = 0 : limδ→0f (x,t+δ)− f (x,t)

δ =d f (x,t)

dt exists by hypothesis. Themean value theorem justifies the following statement:∣∣∣∣ f (x,t+δ)− f (x,t)

δ

∣∣∣∣ =∣∣∣∣d f (x,t

′)

dt

∣∣∣∣ ≤ h(x) for some t′

between t and t + δ. Using a theorem which es-tablishes the continuity of the sought after integral (see [7] page 153) we can conclude thatlimδ→0

F(t+δ)−F(t)δ =

∫X

d f (x,t)dt dµ(x). The Lebesgue Dominated Convergence Theorem ultimately

justifies the theorem which established the continuity of the integral.

14.2.3 Not entirely rigorous but still insightful

The general formula for differentiating under the integral sign, which is called Leibnitz’s Rule,is proved in some calculus textbooks in a non-rigorous (if you are an analyst) but neverthelessinsightful way. One proof [8] goes like this. It may cause some people to fly into a rage becauseit skates over the need for uniform continuity but it nevertheless gives the ”superstructure” forwhy the process works and a good foundation for how to make it rigorous. So with thosequalifications here it is.

Let φ(α) =∫ u2

u1f (x, α) dx where a ≤ α ≤ b and u1, u2 may depend upon α. If f (x, α) and ∂ f

∂α arecontinuous in both x and α in some region of the x, α plane including u1 ≤ x ≤ u2, a ≤ α ≤ band if u1 and u2 are continuous and have continuous derivatives for a ≤ α ≤ b then:

dφ(α)dα

=

∫ u2

u1

∂ f (x, α)∂α

dx + f (u2, α)du2

dα− f (u1, α)

du1

dα(147)

Now if u1 and u2 are constants then we get the simple version of the rule: dφdα =

∫ u2

u1

∂ f (x,α)∂α dx.

To prove this we let φ(α) =∫ u2(α)

u1(α) f (x, α) dx. Then:

∆φ = φ(α + ∆α) − φ(α) =

∫ u2(α+∆α)

u1(α+∆α)f (x, α + ∆α) dx −

∫ u2(α)

u1(α)f (x, α) dx

=

∫ u1(α)

u1(α+∆α)f (x, α+∆α) dx+

∫ u2(α)

u1(α)f (x, α+∆α) dx+

∫ u2(α+∆α)

u2(α)f (x, α+∆α) dx−

∫ u1(α)

u2(α)f (x, α) dx

=

∫ u2(α)

u1(α)[ f (x, α+ ∆α)− f (x, α)] dx +

∫ u2(α+∆α)

u2(α)f (x, α+ ∆α) dx−

∫ u1(α+∆α)

u1(α)f (x, α+ ∆α) dx

(148)

Now the Mean Value Theorem for integrals ensures that:

∃ξ ∈ (α, α + ∆α)∃ξ1 ∈ (u1(α), u1(α + ∆α))∃ξ2 ∈ (u2(α), u2(α + ∆α))

46

such that:∫ u1(α)u2(α) [ f (x, α + ∆α) − f (x, α)] dx = ∆α

∫ u1(α)u2(α)

∂ f (x,ξ)∂α dx∫ u1(α+∆α)

u1(α) f (x, α + ∆α) dx = f (ξ1, α + ∆α)[u1(α + ∆α) − u1(α)]∫ u2(α+∆α)u2(α) f (x, α + ∆α) dx = f (ξ2, α + ∆α)[u2(α + ∆α) − u2(α)]

Therefore, ∆φ∆α =

∫ u2(α)u1(α)

∂ f (x,α)∂α dx + f (ξ2, α + ∆α) ∆u2

∆α − f (ξ1, α + ∆α) ∆u1∆α

Finally letting ∆α→ 0 and using continuity of the derivatives we get the result:dφ(α)

dα =∫ u2(α)

u1(α)∂ f (x,α)∂α dx + f (u2, α) du2

dα − f (u1, α) du1dα . This final step obscures the role of uniform

continuity which is much more prominent in the following line of argument.

14.2.4 Making the Rigour Police happy

To make the Rigour Police happy we have to resort to the quite a bit more machinery. We startwith the following theorem:

Theorem1:

Let fn be a sequence of functions with domainD uniformly converging to f ie fn ⇒ f . If the fnare continuous for each n, then f is continuous onD.

We firstly note what uniform continuity means. The sequence of functions fn converge uniformlyto f on a domain D if for any given ε > 0, there exists an N ∈ N such that , for all x ∈ D,| fn(x)− f (x)| < ε whenever n > N. Note that this is a much stronger requirement than pointwiseconvergence: uniform convergence is a global property while pointwise convergence is a localproperty.

Pointwise convergence amounts to this requirement: If given any ε > 0, for each x ∈ D thereexists a positive integer N(x) ( thus N depends on x) such that | fn(x) − f (x)| < ε whenevern > N(x).

The logical structures of the two definitions are subtly but significantly different. In the caseof pointwise convergence N(x) depends on the choice of x ∈ D (and on ε) while with uniformconvergence the N that is determined may depend on ε but cannot depend on x ∈ D. Therequirement, ”there exists an N ∈ N such that , for all x ∈ D” etc makes this clear.

If we consider the functions defined as follows:

gn(x) =nx

nx + 112≤ x ≤ 1 (149)

we can show that the gn converge pointwise to the limit g(x) = 1 for 12 ≤ x ≤ 1. Thus |gn(x) −

g(x)| = | nxnx+1 − 1| = 1

nx+1 ≤1nx < ε if we choose N(x) = 1 +

[ 1εx

]. For instance, suppose ε = 0.1

and x = 12 then N( 1

2 ) = 21 so that |g21( 12 ) − g( 1

2 )| = 1212 +1

= 223 <

110 = ε.

47

To show that the gn(x) converge uniformly to g(x) = 1 on [ 12 , 1] we note that |gn(x) − g(x)| =

1nx+1 <

1nx = ε

n1εx ≤

εn

1ε 2 using the fact that x ≥ 1

2 and so 1x ≤ 2. Thus if we choose n > 2

ε (notethat this is independent of x) we will have 1

nx <ε2

1x <

ε2 2 = ε and so the convergence is uniform

on [ 12 , 1] . See the diagram below:

With this background we can now prove Theorem 1 as follows. Let x0 ∈ D and xn be asequence in D such that xn → x0. We use sequences because we know that f is continuous atx0 if and only if f (xn) → f (x0) when xn → x0. So we choose our ε > 0 and note that becausefn(x) ⇒ f (x) there exists a positive integer N1 such that | fn(xm) − f (xm)| < ε

3 for all m ∈ Nand all n > N1. Note here that the requirement m ∈ N is merely picking up the fact that in thedefinition of uniform continuity the requirements hold for any x ∈ D and so must hold for allelements of the sequence. Moreover, N1 works for x0 and all xm because that is what uniformcontinuity means.

Because fn is continuous at x0 we can find an N such that | fn(xm) − fn(x0)| < ε3 for all m >

N

We now do the standard trick of splitting up the relevant absolute difference (ie | f (xm) − f (x0)|)into, in this case, 3 separate absolute differences which we can separately make arbitrarily small.Thus:

48

| f (xm) − f (x0)| = | f (xm) − fn(xm) + fn(xm) − fn(x0) + fn(x0) − f (x0)|

≤ | f (xm) − fn(xm)| + | fn(xm) − fn(x0)| + | fn(x0) − f (x0)| <ε

3+ε

3+ε

3= ε (150)

for all m > maxN1,N.

This shows that f (xm)→ f (x0) which means that f is continuous at x0 and since x0 was arbitrary,at all points inD.

We need another theorem as a foundation for the ultimate proof.

Theorem2:

Suppose fn is a sequence of functions all of which are integrable on [a, b] and uniformlyconvergent with limit f . For all n ∈ N, let gn(x) =

∫ xa fn(t) dt for x ∈ [a, b]. Then gn(x)

also converges uniformly on [a, b] and limn→∞ gn(x) = g(x) where g(x) =∫ x

a f (t) dt. Inother words, we can interchange the limiting process with integration ie limn→∞

∫ xa fn(t) dt =∫ x

x limn→∞ fn(t) dt.

To prove this theorem we take, as usual, ε > 0 and note that because the fn uniformly convergeto f there exists an N such that | fn(x) − f (x)| < ε

b−a for all x ∈ [a, b] and for all n > N. Becausewe want to show that gn(x) converges uniformly to g(x) on [a, b] we consider the difference:

|gn(x) − g(x)| = |∫ x

afn(t) dt −

∫ x

af (t) dt| = |

∫ x

a

(fn(t) − f (t)

)dt|

≤

∫ x

a| fn(t) − f (t)| dt <

ε

b − a

∫ x

adt <

ε

b − a(b − a) = ε (151)

Now because (151) holds for all x ∈ [a, b] provided n > N where N is independent of x thismeans that gn ⇒ g on [a, b].

Theorem3:

Let fn be a sequences of functions which are differentiable on [a, b] such that fn → f (iepointwise convergence). If the derivatives f

′

n are continuous on [a, b] for all n ∈ N and ifthe sequence f

′

n is uniformly convergent, then limn→∞ f′

n = f′

. Thus the limit of uniformlyconvergent derivatives is the derivative of the limit.

To prove this theorem we first note that since f′

n is uniformly convergent there must be somefunction h such that f

′

n ⇒ h. By Theorem 1 h is continuous on [a, b]. Now let gn(x) =∫ xa f

′

n(t) dt = fn(x) − fn(a) for x ∈ [a, b] and g(x) =∫ x

a h(t) dt for x ∈ [a, b]. The hypothe-ses of Theorem 2 are satisfied (note that the f

′

n are integrable on [a, b]) so that gn → g. But forevery x ∈ [a, b], gn(x) = fn(x) − fn(a) → f (x) − f (a) (this is because of the assumed pointwiseconvergence of f ). Hence g(x) = f (x) − f (a) for all x and this limit is unique. By applyingthe Fundamental Theorem of Calculus we see that since h is continuous, g is differentiable on

49

[a, b] and g′

(x) = h(x). So f is also differentiable on [a, b] and g′

= f′

= h which finalises theproof.

Theorem4:

Let g : R × R → C be a continuous function such that ∂g(x,t)∂t exists and is continuous. Then∫ b

a g(x, t) dx is differentiable with ddt

∫ ba g(x, t) dx =

∫ ba

∂g(x,t)∂t dx

To prove this we need to note that since ∂g(x,t)∂t is continuous we can interchange the order of

integration in the following integrals [see [3], pages 291-297]:

∫ y

0

( ∫ b

a

∂g(x, t)∂t

dx)

dt =

∫ b

a

( ∫ y

0

∂g(x, t)∂t

dt)

dx =

∫ b

a

(g(x, y) − g(x, 0)

)dx

=

∫ b

ag(x, y) dx −

∫ b

ag(x, 0) dx (152)

Therefore:

∫ b

ag(x, y) dx =

∫ y

0G(t) dt + constant (153)

where G(t) =∫ b

a∂g(x,t)∂t dx

Recall that if G is continuous at each y ∈ R then ddy

( ∫ y0 G(t) dt+constant

)= G(y) =

∫ ba

∂g(x,y)∂y dx.

This flows from the fact that continuity of the partial derivative ensures integrability which inturn ensures continuity of the integral. The Fundamental Theorem of Calculus lets us differenti-ate the integral.

Because ∂g(x,t)∂t is continuous on each closed subset of R × R it is uniformly continuous on all

such subsets. This allows us to perform the following estimate:

|G(t) −G(y)| =∣∣∣∣ ∫ b

a

∂g(x, t)∂t

dx −∫ b

a

∂g(x, y)∂y

dx∣∣∣∣ =

∣∣∣∣ ∫ b

a

(∂g(x, t)∂t

−∂g(x, y)∂y

)dx

∣∣∣∣≤

∫ b

a

∣∣∣∣∂g(x, t)∂t

−∂g(x, y)∂y

∣∣∣∣ dx ≤ (b − a) ε (154)

where we have used the fact that the partial derivative is uniformly continuous.

This establishes that G is continuous and so∫ b

a g(x, y) dx is differentiable. Thus:

ddy

∫ b

ag(x, y) dx = G(y) =

∫ b

a

∂g(x, y)∂y

dx (155)

50

14.2.5 Proof of the Main Result

Let fn(t) =∫ n−n g(x, t) dx and f (t) =

∫ ∞−∞

g(x, t) dx. Then:

| fn(t) − f (t)| =∣∣∣∣ ∫ n−n g(x, t) dx −

∫ ∞−∞

g(x, t) dx∣∣∣∣ =

∣∣∣∣ ∫|x|>n g(x, t) dx∣∣∣∣

≤∫|x|>n

∣∣∣∣g(x, t)∣∣∣∣ dx → 0 since by assumption

∫ ∞−∞|g(x, t)| dx exists and so the tails must approach

0 for sufficiently large n. Thus fn → f pointwise.

Now fn is differentiable for all n where f′

n(t) = ddt

∫ n−n g(x, t) dx =

∫ n−n

∂g(x,t)∂t dx using Theorem 4.

Furthermore∣∣∣∣ f ′n(t) −

∫ ∞−∞

∂g(x,t)∂t dx

∣∣∣∣ =∣∣∣∣ ∫ n−n

∂g(x,t)∂t dx −

∫ ∞−∞

∂g(x,t)∂t dx

∣∣∣∣≤

∫|x|>n

∣∣∣∣∂g(x,t)∂t

∣∣∣∣ dx→ 0 uniformly on each interval [a, b] by assumption.

Thus f (t) =∫ ∞−∞

g(x, t) dx is differentiable and ddt

∫ ∞−∞

g(x, t) dx =∫ ∞−∞

∂g(x,t)∂t dx

For more reading on differentiating under the integral sign see [9] and [10].

15 Proof that∫ ∞

0sin x

x dx = π2

This integral was pivotal to Dirichlet’s work on Fourier series in the 19th century. There are var-ious ways to prove it and Dirichlet’s original proof is referred to in [18] at page 446. Churchill’sproof ( [17] pages 85-86) is relatively straightforward and runs as follows.

We first note some basic properties of S (x) = sin xx for x , 0 and S (0) = 1.

S (x) = 1 −x2

3!+

x4

5!−

x6

7!+ . . . (156)

S (x) is even and the series in (156) is an alternating one. Because we can write S (x) = 1− ( x2

3! −x4

5! )− ( x6

7! −x8

9! ) + . . . , for 0 < x ≤ 1 it follows that 0 < S (x) < 1. When x > 1 then |S (x)| < |sin x|so we can say that |S (x)| ≤ 1 for all x ≥ 0.

Now to get an estimate for the integral we simply break the domain of integration up into inter-vals of length π as follows: (0, π), (π, 2π), . . . and note that the sign of S (x) alternates on eachsuccessive interval. Thus if we choose ν > 0 and n as the greatest integer such that nπ ≤ ν wewill have:

∫ ν

0S (x) dx =

∫ π

0S (x) dx +

∫ 2π

πS (x) dx + · · · +

∫ nπ

(n−1)πS (x) dx +

∫ ν

nπS (x) dx (157)

Now if we let Ak =∣∣∣ ∫ (k+1)π

kπ S (x) dx∣∣∣ we can write (157) in the following form:

51

∫ ν

0S (x) dx = A0 − A1 + A2 − · · · + (−1)n−1 An−1 + (−1)nθn An (158)

Note that θn ∈ [0, 1). Now A0 < π because A0 = |∫ π

0 S (x) dx| ≤∫ π

0 |S (x)| dx < π. On the kth

interval (k > 0) we have that |S (x)| < 1kπ so that Ak < π 1

kπ = 1k . We thus see that Ak+1 < Ak.

Because the Ak constitute a sequence of positive terms which converge to zero as k → ∞ thealternating series limn→∞

∑n−1k=0 (−1)k Ak converges.

The improper integral in (158) exists as ν→ ∞ since θn An → 0 as n→ ∞.

If we write:

F(t) =

∫ ∞

0e−tx sin x

xdx for t ≥ 0 (159)

we can be sure that the integral in (159) converges because |F(t)| ≤∫ ∞

0 e−tx | sin xx | dx ≤

∫ ∞0 e−tx dx =

1t

We can write (158) as follows because the integral exists for t ≥ 0:

F(t) = limn→∞

n−1∑k=0

(−1)k∫ (k+1)π

kπe−tx sin x

xdx (160)

We know from the discussion following (145) that if we take the absolute value of F(t) in (160)we will have for every n, |F(t)| ≤

∑n−1k=0

∫ (k+1)πkπ

∣∣∣ sin xx

∣∣∣ dx, and the absolute value of the remainderterm An ≤

1n which is independent of t. This gives us uniform continuity of the series and hence

continuity of F(t). Thus F(+0) = F(0) =∫ ∞

0sin x

x dx.

We can get an expression for F′

(t) by differentiating under the integral sign (which is kosher -see the earlier discussion in the Appendix) and then integrating by parts.

Thus:

F′

(t) = −

∫ ∞

0e−tx sin x dx (161)

Now integrating (161) by parts we get:

−

∫ ∞

0e−tx sin x dx =

[e−tx cos x

]∞0

+ t∫ ∞

0e−tx cos x dx = −1 + t

∫ ∞

0e−tx cos x dx

= −1 + t[

e−tx sin x]∞0

+ t∫ ∞

0e−tx sin x dx

= −1 + t2

∫ ∞

0e−tx sin x dx (162)

52

Therefore:

(1 + t2)∫ ∞

0e−tx sin x dx = −1 =⇒ F

′

(t) =−1

1 + t2 (163)

From (163) it follows that:

F(t) = − arctan t + C (164)

Because |F(t)| ≤ 1t it follows that F(t) → 0 as t → ∞. Thus C = π

2 because arctan t → π2 as

t → ∞. Finally, because F(0) =∫ ∞

0sin x

x dx we have the result:∫ ∞

0sin x

x dx = π2 .

53