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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}

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