# Welcome.

## January 12, 2011

### Harmonic Analysis Lecture Notes 5

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:53 pm

Note to self: this set of notes requires some fixing.

Theorem (Carathéodory–Toeplitz/Bochner) (CT for ${{\mathbb R}/{\mathbb Z}}$, B for ${{\mathbb R}^n/{\mathbb Z}^n}$) If ${d\mu}$ is a positive measure on ${{\mathbb R}^d}$ then

$\displaystyle (\text{something missing here})\left[\frac{\hat mu(0) \hat mu(\xi_2 -\xi_1)}{\hat u(\xi_2 - \xi_1)}\right] \geq 0$

for all ${\xi_1, \ldots, x_N \in {\mathbb R}^d}$ and all ${N}$. Conversely if ${\hat mu}$ is a (hermitian symmetric) continuous function on ${{\mathbb R}^d}$ then ${\hat mu}$ is the Fourier transform of a positive measure when all matrices are positive semi-definite

Proof: If ${d\mu \geq 0}$ then for all ${c_1, \ldots, c_N \in {\mathbb C}}$ and all ${\xi_1, \ldots, \xi_N \in {\mathbb R}^d}$

$\displaystyle \begin{array}{rcl} 0 \leq \int |\sum c_n e^{-2\pi i \xi \cdot x}|^2 \; d\mu &=& \displaystyle \sum_{n,m} \bar c_m c_n \int e^{-2\pi i(\xi_n - \xi_m) \cdot x} \; d\mu \\ &=& \displaystyle \sum_{n,m} \bar c_m c_n \hat \mu(\xi_n - \xi_m) \end{array}$

I.e., the matrix with entries ${\hat u (\xi_n - \xi_m)}$ is positive semi-definite. (Not that since ${\mu}$ is a real measure, ${\hat \mu(-\xi) = \overline{ \hat u (\xi)}}$ so the matrix is Hermitian).

For the converse, we use an idea of Féjer. By taking Riemann sums

$\displaystyle \begin{array}{rcl} 0 &\leq& \int_0^h \int_0^h \hat \mu(t - s)e^{2\pi i ty} e^{-2\pi i sy} \; ds \; dt \\ \\ &=& \int \Big| \int_0^h e^{-2\pi it(x-y)} \; dt \Big|^2 d\mu(x) \\ \\ &=& \int_0^h \int_0^h \int e^{-2\pi i(t - s) (x-y)} \; d\mu(x) \; dt \; ds \end{array}$

and so if we define

$\displaystyle f_h(y) := h \int_0^h \int_0^h e^{-2\pi i(s - t) y} \hat \mu(t - s) \; ds \; dt$

then ${f_h(y) \geq 0}$ because

$\displaystyle \begin{array}{rcl} \displaystyle h \int_0^h \int_0^h e^{-2\pi i(s - t) y} \hat mu(t - s) \; ds \; dt &=& h \displaystyle \int \Big| \frac{1- e^{-2\pi i h(x-y)}}{2\pi i h(x-y)} \Big|^2 \; d\mu(x) \\ \\ &=& \displaystyle \int \frac{\sin^2(\pi h(x-y))}{\pi^2 (x-y)^2 h} \; d\mu(x) \end{array}$

Now observe that positive measure ${f_h(x) \; dx}$ converges weak-${*}$ to ${c d\mu}$ for some ${0 < c < \infty}$ (uniqueness of weak-${*}$ limit points follows from ${\hat f \rightarrow c \hat u}$). Now,

$\displaystyle \int \varphi(x) f_h(x) \; d\mu(x) = \int \int \frac{\sin^2(\pi h(x-y))}{\pi^2(x-y)h} \varphi(y) \; d\mu(x).$

which converges as ${n \rightarrow \infty}$ to c ${\varphi (x)}$ where ${c = \int \frac{\sin^2(\pi x)}{\pi^2 x^2} \; dx}$. $\Box$

We can do better than

$\displaystyle f(x) = \lim_{h \rightarrow \infty} \int_{-h}^h e^{-2\pi i x \cdot \xi} \hat f(\xi) \; d\xi$

via

$\displaystyle f(x) = \lim_{h\rightarrow\infty} \int_{-h}^h \prod_{j} \left( 1- \frac{|\xi_h|}{h}\right)e^{-2\pi i x \cdot \xi} \hat f(\xi) \; d\xi$

Note that ${\chi_I * \chi_I = 1 - \frac{|\xi|}{h}}$. It follows immediately from the following theorem that the Fourier transform extends continuously from ${\mathcal{S}({\mathbb R}^d)}$ to ${L^p({\mathbb R}^d)}$ when ${1 \leq p \leq 2}$, yielding a function in a (thin subset) of ${L^{p'}({\mathbb R}^d)}$, where ${\frac{1}{p} + \frac{1}{p'} = 1}$.

Theorem (Hausdorff-Young) For all ${f \in \mathcal{S}({\mathbb R}^d)}$,

$\displaystyle ||\hat f ||_{p'} \leq ||f||_{p}$

where ${1 \leq p \leq 2}$ and ${\frac{1}{p} + \frac{1}{p'} = 1}$.

Proof: This assertion is true when ${p=1}$ by Minkowski’s inequality and when ${p = 2}$ by Plancherel’s theorem. The result follows from the Riesz–Thorin interpolation theorem (cf below). $\Box$

Theorem (Riesz–Thorin Interpolation) Suppose ${T}$ is a ${{\mathbb C}}$-linear transformation and

$\displaystyle ||Tf||_{q_j} \leq M_j ||f||_{p_j}$

with ${1 \leq p_j, q_j \leq \infty}$ and ${j = 0,1}$. Then

$\displaystyle ||Tf||_{q_\theta} \leq M_0^{1-\theta} M_1^{\theta} ||f||_{p_\theta}$

where

$\displaystyle \frac{1}{p_\theta} = \frac{1-\theta}{p_0} + \frac{\theta}{p_1} \text{\;\;\; and \;\;\;} \frac{1}{q_\theta} = \frac{1-\theta}{q_0} + \frac{\theta}{q_1}$