Math 6338 : Real Analysis II

Final ExamDue: Friday May 4 2012 at 5:00pm

Instructions: Answer all of the problems. You may use course notes, but other sources arenot permitted.

1. Let H be a complex separable Hilbert space. A linear operator T : H H is completelycontinuous if any weakly convergent sequence {xn} is mapped to a strongly convergentsequence {Txn}.

(a) Show that T is completely continuous if and only if T is compact.

(b) Suppose that T T is completely continuous. Show that T is completely continuous.

Solution: Part (a): If T is compact, then weakly convergent sequences are mapped tostrongly convergent sequences, as was proved in class. This gives that T is completelycontinuous.

Suppose now that T is completely continuous. Since H is separable, the unit ball ofH is weakly compact. Let {xn} be any sequence in the unit ball of H, then there is ax H and a subsequence {xnk} such that xnk x weakly. But since T is completelycontinuous we have that

Txnk TxH 0.This implies that the image of the unit ball of H is sequentially compact, and so T is acompact operator.

Part (b): Let {xn} be a weakly convergent sequence in H. Since it is weakly convergent,it is in fact bounded and so xnH C for all n. Because T T is completely continuouswe have that {T Txn} is strongly convergent. We now claim that {Txn} is a Cauchysequence. Indeed, since

Txn Txm2H = T (xn xm), T (xn xm)H= T T (xn xm), xn xmH T T (xn xm)H xn xmH . 2C T Txn T TxmH .

This last expression can be made small if n,m are large enough, and so we have that{Txn} is Cauchy in H, and hence converges.

2. Let 1 < p < and 1p

+ 1q

= 1. Suppose that K(x, y) : Rn Rn R is measurable andK(x, y) is non-negative almost everywhere. If

supxRn

RnK(x, y) dy A and sup

yRn

RnK(x, y) dx B

show that Tf(x) =Rn K(x, y)f(y) dy is a bounded operator on L

p(Rn) and

TfLp(Rn) A1qB

1p fLp(Rn) .

Solution: Note that we have

TfpLp(Rn) =Rn

RnK(x, y)f(y)dy

p dx

Rn

Rn

(RnK(x, y) |f(y)| dy

)pdx

=

Rn

Rn

(RnK(x, y)

1qK(x, y)

1p |f(y)| dy

)pdx

Rn

(RnK(x, y)dy

) pq(

RnK(x, y) |f(y)|p

)dx

A pqRn

RnK(x, y) |f(y)|p dydx

= Apq

Rn|f(y)|p

(RnK(x, y)dx

)dy

A pqB fpLp(Rn) .

Taking the p-th root gives the result.

3. Suppose that L is a linear functional on C[0, 1] such that

L = L(1) = 1.Let f C[0, 1] with 0 f 1. Then show that 0 Lf 1.

Solution: Let f C[0, 1] with 0 f 1. Set Lf = + i. We want to show that = 0 and 0 1. For all t R we have that

L

(f 1

2+ it

)= 1

2+ i( + t).

We have thatf 1

2

12 , we have that(

12

)2+ ( + t)2 =

L(f 12 + it)2 f 12

2 1

4+ t2.

Rearrangement gives that

2 + 2 + 2t 0 t R.

So = 0, and hence 2 , or rephrasing, 0 1 or 0 Lf 1.

4. Let X and Y be Banach spaces and let B(, ) be separately continuous bilinear mappingX Y to C. Namely, for fixed x, the mapping y B(x, y) is a bounded linear transforma-tion, and with a similar statement holding for fixed y. Show that B(, ) is jointly continuous.Namely xn 0 and yn 0, then B(xn, yn) 0.

Solution: Set Tn(y) = B(xn, y). Since for fixed xn the mapping B(xn, ) is continuous,each Tn : Y C is bounded. Because xn 0 and B(, y) is bounded, we have thatfor all y Y that {Tn(y)} is bounded for each fixed y. By the Uniform BoundednessPrinciple, there is a constant C such that

Tn(y) C y n.

Then we have that|B(xn, yy)| = Tn(yn) C yn 0.

5. Let f S(R).(a) Show that f2L2(R) 4pi xfL2(R)

fL2(R)

.

(b) Show that equality holds in this inequality if and only if f(x) = Ceax2

where a > 0and C C.

Solution: Part (a): Note that by integration by parts and direct computation we have

f2L2(R) =R|f(x)|2 dx =

Rxd

dx|f(x)|2 dx = 2Re

(Rf(x)f (x)dx

).

Then, applying Cauchy-Schwarz, we have

f2L2(R) 2R|x| |f(x)| |f (x)| dx

2 xfL2(R) f L2(R) .

Now apply Plancherels Theorem to conclude that

f L2(R) =f

L2(R)= 2pi

fL2(R)

,

where for the last equality we have used that f () = 2piif().

Part (b): In the application of Cauchy-Schwarz we have an equality if and only if

2axf(x) = f (x).

Solving we have f(x) = Ceax2

for some C C and a R. The choice of a > 0 isrequired for f L2(R).