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En la frontera del Analisis Armonico con laTeorıa de los Numeros

Congreso Centenario RSME5 de febrero de 2011

Antonio Cordoba (UAM) Harmonic Analysis and Number Theory Centenario RSME 1 / 31

[. . . ] the usefulness of Fourier series is not

limited to research in Physics; they have been

successfully applied also to a field in pure

mathematics, namely Number Theory, and here it

seems to be of importance precisely to consider those

functions whose representability by trigonometric

series has not yet been investigated [. . . ]

Bernhard Riemann

Harmonic Analysis

On euclidean space Rn, Tn (periodic functions).

On the hyperbolic plane, SL2(R), automorphic functions and forms.

f (ν) =

e−2πiν·x f (x)dx , ν ∈ Zn

f (ξ) =

e−2πiξ·x f (x)dx , ξ ∈ Rn

Questions:

lımΩ→Rn

∑ν∈Ω

f (ν)e2πiν·x = f (x) ?

lımΩ→Rn

f (ξ)e2πiξ·xdx = f (x) ?

Harmonic Analysis

Important identities:

On Tn: ∫Tn

|f (x)|2dx =∑ν∈Zn

|f (ν)|2 (Bessel)

On Rn: ∑ν∈Zn

f (ν) =∑ν∈Zn

f (ν) (Poisson)

On SL2(R):“trace formula” (Selberg)

Analisis Armonico

Verde, verde esmeralda,azul turquesa, azul ultramar,ındigo, violeta:sıntesis de luz.

Ondas, vibraciones, trigonometrıa.Espirales, remolinos, puntos de fuga.Venus de proporciones divinas.

Fuego que da la vida,el calor y el color.Amarillo, naranja,rojo, carmın.

The value of ζ(2)

Grafitti valenciano

The value of ζ(2)

Fractionary part:

x −m, |x −m| < 1/2;0, x = m + 1/2.

Its Fourier series:

∞∑n=1

(−1)n+1 1

nsin(2πnx)

Bessel’s identity yields:

0x2dx =

∞∑n=1

n2=⇒ ζ(2) =

The value of ζ(2)

A new proof:

ζ(2) =∞∑

n2=∞∑

(2k)2+∞∑

(2k + 1)2=ζ(2)

4+∞∑

(2k + 1)2

ζ(2) =4

∞∑k=0

(2k + 1)2=

∞∑k=0

(∫ 1

0x2kdx

)(∫ 1

0y2kdy

( ∞∑k=0

x2ky2k

)dx dy =

1− x2 y2

The value of ζ(2)

Change of variables: hyperbolic tangent.

x = tanh(u) =

eu − e−u

eu + e−u

y = tanh(v) =ev − e−v

ev + e−v

cosh2(u)du

cosh2(v)dv

The value of ζ(2)

Therefore,

ζ(2) =1

∫ ∞−∞

cosh(u) cosh(v)− sinh(u) sinh(v)

cosh(u) cosh(v) + sinh(u) sinh(v)dudv

∫ ∞−∞

cosh(u − v) cosh(u + v)

The value of ζ(2)

The change of variables s = u − v , t = u + v yields

ζ(2) =1

∫ ∞−∞

cosh(s)

∫ ∞−∞

cosh(t)=

[∫ ∞−∞

cosh(s)

because ∫ ∞−∞

cosh(s)=

∫ ∞−∞

es + e−s

∫ ∞0

1 + z2= π .

Analyst = hunter

Hunting team:

J. Cilleruelo C. FeffermanF. Chamizo D. CordobaP. Balodis F. GancedoP. Fernandez A. Ubis

C. Vinuesa...

Analyst = hunter

Hunting team:

C. Vinuesa...

Analyst = hunter

Hunting team:

C. Vinuesa...

Sharpening the tools: Singular integrals

Tf (x) = K ∗ f (x) = lımε→0

∫|x−y |>ε

Ω(x − y)

|x − y |nf (y) dy

• Ω(λx) = Ω(x) , λ > 0, x ∈ Sn−1 (homogeneous of degree 0);

•∫

Sn−1

Ω(x)dσ(x) = 0 (mean value 0);

The multiplier:

Tf (ξ) = K (ξ) f (ξ) , m(ξ) = K (ξ) ∈ L∞(Rn).

Key fact:

T ∼ identity : ‖Tf ‖p ≤ Cp‖f ‖p

Riesz transforms:

Rjk f (x) = p.v.

(xj − yj)(xk − yk)

|x − y |n+2f (y)dy ,

∂xj∂xk= Rjk(∆f ).

The multiplier:

Tf (ξ) = K (ξ) f (ξ) , m(ξ) = K (ξ) ∈ L∞(Rn).

Key fact:

Riesz transforms:

Rjk f (x) = p.v.

|x − y |n+2f (y)dy ,

The multiplier:

Tf (ξ) = K (ξ) f (ξ) , m(ξ) = K (ξ) ∈ L∞(Rn).

Key fact:

Riesz transforms:

Rjk f (x) = p.v.

|x − y |n+2f (y)dy ,

Calderon-Zygmund

For f ∈ L1(Rn) and α > 0, there are Qj disjoint cubes such that

• α ≤ 1

|f (y)|dy ≤ 2nα

• |f (x)| ≤ α a.e. in Rn \ ∪jQj

Thenf (x) = g(x) + b(x)

b(x) =

f (x)− 1

|Qj |∫Qj|f (y)|dy , x ∈ Qj

0, x ∈ Rn \ ∪jQj .

Piet Mondrian

Bochner-Riesz

Tαf (ξ) = (1− |ξ|2)α+ f (ξ) , Tαf (x) = (Kα ∗ f )(x) (α ≥ 0)

The kernel verifies

Kα(ξ) = (1− |ξ|2)α+ , Kα(x) ≈ e i |x |

|x |n+1

, x →∞

Theorem

Tα is bounded on Lp(R2), for 4/(3 + 2α) < p < 4/(1− 2α).

T0 is only bounded on L2(R2).

Bochner-Riesz

The kernel verifies

Kα(ξ) = (1− |ξ|2)α+ , Kα(x) ≈ e i |x |

|x |n+1

, x →∞

Theorem

Bochner-Riesz

The kernel verifies

Kα(ξ) = (1− |ξ|2)α+ , Kα(x) ≈ e i |x |

|x |n+1

, x →∞

Theorem

Bochner-Riesz

Kazimir Malevich

Lattice point Problems

Letr(n) = #(j , k) ∈ Z2|n = j2 + k2.

First cases:

r(1) = 4, 1 = 02 + (±1)2 = (±1)2 + 02

r(3) = 0

r(5) = 8, 5 = (±1)2 + (±2)2 = (±2)2 + (±1)2

r(7) = 0...

• lımn→∞

nε= 0 , ∀ε > 0.

• lım supn→∞

(ln n)a=∞ , ∀a > 0.

Let us define:

R(x) =∑

m≤x2

r(n) =∑ν∈Z2

χDx(ν) , χDx

(ν) =

1, ν ∈ Dx

0, ν /∈ Dx

ThenR(x) = πx2 + E (x) .

Gauss: E (x) = O(x)

R(x) =∑ν

χDx(ν) =

χDx(ν) = πx2 +

∑|ν|≥1

χDx(ν)

︸︷︷︸=E(x)

Let us define:

R(x) =∑

m≤x2

r(n) =∑ν∈Z2

χDx(ν) , χDx

(ν) =

1, ν ∈ Dx

0, ν /∈ Dx

Gauss: E (x) = O(x)

R(x) =∑ν

χDx(ν) =

χDx(ν) = πx2 +

∑|ν|≥1

χDx(ν)

︸︷︷︸=E(x)

Let us define:

R(x) =∑

m≤x2

r(n) =∑ν∈Z2

χDx(ν) , χDx

(ν) =

1, ν ∈ Dx

0, ν /∈ Dx

Gauss: E (x) = O(x)

R(x) =∑ν

χDx(ν) =

χDx(ν) = πx2 +

∑|ν|≥1

χDx(ν)

︸︷︷︸=E(x)

E (x) = O(x2/3) (G. Voronoi, W. Sierpinski, 1903)

· · · (J. Van der Corput, L. Hua, G. Kolesnik. . . )

E (x) = O(x131/208) (H. Iwaniec-C. Mozzochi, 1993; M. Huxley, 2001)

Recall that 23 = 0,666 . . . and 131

208 = 0,6298 . . . .

On the other direction:

lım supx→∞

x1/2=∞ (G. Hardy-Landau).

Conjecture

E (x) = O(x1/2+ε), for all ε > 0.

Recall that 23 = 0,666 . . . and 131

208 = 0,6298 . . . .

lım supx→∞

Conjecture

E (x) = O(x1/2+ε), for all ε > 0.

Recall that 23 = 0,666 . . . and 131

208 = 0,6298 . . . .

lım supx→∞

Conjecture

E (x) = O(x1/2+ε), for all ε > 0.

Lattice points on “small arcs”

Theorem (J. Cilleruelo, A.C.)

For each α < 1/2, there exists a constant Cα <∞ such that every arc oflength Rα (on a circle centered at 0 and radius R) contains, at most, Cαlattice points.

Problem. What happens when 12 ≤ α < 1 ?

Lattice points on “small arcs”

Theorem (J. Cilleruelo, A.C.)

For each α < 1/2, there exists a constant Cα <∞ such that every arc oflength Rα (on a circle centered at 0 and radius R) contains, at most, Cαlattice points.

Problem. What happens when 12 ≤ α < 1 ?

Path integrals and lattice points

Take a function f (x) such that f (0) = f (1) = 0 and0 < c1 ≤ −f ′′(x) ≤ c2. F (x) will be the rescaling:

ThenF (0) = F (N) = 0 , 0 <

N< −F ′′(x) <

Let us call yj = F (j):

[yj ] = #lattice points in the vertical segment [(j , 0), (j , yj)]

Take a function f (x) such that f (0) = f (1) = 0 and0 < c1 ≤ −f ′′(x) ≤ c2. F (x) will be the rescaling:

ThenF (0) = F (N) = 0 , 0 <

N< −F ′′(x) <

Let us call yj = F (j):

[yj ] = #lattice points in the vertical segment [(j , 0), (j , yj)]

Define

EN(f ) =

0F (x)dx −

N∑j=0

([F (j)] +

Theorem (F. Chamizo, A.C.)(∫EN(f )2dµ(f )

)1/2= O(N1/2+ε) for all ε > 0.

Rudin’s conjecture

f ∈ L1(T) , f ∼∑n

ane2πin2x =⇒ f ∈ Lp(T) , ∀p < 4 ?

A related question:

How many square integers are there in an arithmetical progression oflength N?

Rudin’s conjecture implies

#(a + nr |n = 0, 1, . . . ,N − 1 ∩ m2

)= O(N1/2+ε) , ∀ε > 0.

Proposition (A.C.)

Rudin’s conjecture holds under the “tauberian condition”:a0 ≥ a1 ≥ a2 ≥ · · ·

f ∈ L1(T) , f ∼∑n

ane2πin2x =⇒ f ∈ Lp(T) , ∀p < 4 ?

A related question:

#(a + nr |n = 0, 1, . . . ,N − 1 ∩ m2

)= O(N1/2+ε) , ∀ε > 0.

Proposition (A.C.)

f ∈ L1(T) , f ∼∑n

ane2πin2x =⇒ f ∈ Lp(T) , ∀p < 4 ?

A related question:

#(a + nr |n = 0, 1, . . . ,N − 1 ∩ m2

)= O(N1/2+ε) , ∀ε > 0.

Proposition (A.C.)

f ∈ L1(T) , f ∼∑n

ane2πin2x =⇒ f ∈ Lp(T) , ∀p < 4 ?

A related question:

#(a + nr |n = 0, 1, . . . ,N − 1 ∩ m2

)= O(N1/2+ε) , ∀ε > 0.

Proposition (A.C.)

The corresponding multiplier:

TN f (x) =N−1∑k=0

f (k2) e2πik2x .

Question:

‖TN f ‖p ≤ Cp‖f ‖2 , 2 ≤ p ≤ 4, uniformly on N?

Recall that

TN f = GN ∗ f with GN(x) =N−1∑k=0

e2πik2x (Gaussian sum).

Riemann’s series

Let us consider, for δ > 1,

∞∑n=1

e2πinkx

nδ=∞∑

cos(2πnkx)

nδ+ i

∞∑n=1

sin(2πnkx)

nδ= Fk,δ(x) + i Gk,δ(x).

Theorem (F. Chamizo, A.C.)

Fk,δ and Gk,δ are fractals whose Minkowski dimension is 2 + 12k −

Riemann’s series

Let us consider, for δ > 1,

∞∑n=1

e2πinkx

nδ=∞∑

cos(2πnkx)

nδ+ i

∞∑n=1

sin(2πnkx)

nδ= Fk,δ(x) + i Gk,δ(x).

Theorem (F. Chamizo, A.C.)

Fk,δ and Gk,δ are fractals whose Minkowski dimension is 2 + 12k −

Riemann’s series

Young hunters

+ −→ Generalized Sidon sets(Cilleruelo, Vinuesa and Ruzsa)

+ −→ Ergodic theorems in SL2(Z)/SL2(R)(Ubis, Sarnak)

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