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January 14, 2011

Harmonic Analysis Lecture Notes 6

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

Recall from the previous lecture

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

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

Proof: It suffices to bound

$\displaystyle \int \bar g (Tf) \; dx \leq M_0^{1 - \theta} M_1^\theta ||f||_{p_\theta} ||g||_{q'_\theta}$

Note ${\frac{1}{q_\theta'} + \frac{1}{q_\theta} = 1}$ and ${\hat \; '}$ commutes with ${-\theta}$; in particular, ${(q')_\theta = (q_\theta)'}$. Setting

$\displaystyle \varphi(z) = \int \frac{\hat g}{|g|} |g|^{\left( \frac{1-z}{q_0'} + \frac{z}{q_1'}\right) q_\theta'} T\left( \frac{f}{|f|} |f|^{\left( \frac{1-z}{p_0} + \frac{z}{p_1}\right) p_\theta} \right) \; dx,$

then

1. ${\int \hat g Tf = \varphi(\theta)}$
2. We have the inequality

$\displaystyle \begin{array}{rcl} \varphi(0 + it) &\leq& \displaystyle ||g^{\left( \frac{1-z}{q_0'} + \frac{z}{q_1'}\right) q_\theta'}||_{q_0'} ||T||_{L^{p_0} \rightarrow L^{q_0}} || f^{\left( \frac{1-z}{p_0} + \frac{z}{p_1}\right) p_\theta} ||_{p_0} \\ \\ &\leq& || \; |g|^\frac{q_\theta'}{q_0'} \; ||_{q_0'} M_0 || \; |f|^\frac{p_\theta}{p_0}\; ||_{p_0} \end{array}$

and hence if ${||g||_{q_\theta'} = ||f||_{p_\theta'} = 1}$, then

$\displaystyle \varphi(0 + it) \leq M_0$

and similarly

$\displaystyle \varphi(1 + it) \leq M_1$

3. ${\varphi}$ is a bounded holomorphic function in a neighborhood of ${0 \leq \text{Re } z \leq 1}$.

Recalling the Hadamard three lines theorem (Phragmen–Lindelöf principle), we obtain the result (${\log|\varphi(z)|}$ is sub-harmonic, sub-harmonic functions obey the maximum principle, the fact that ${\varphi}$ is bounded means that the part of the boundary at infinity does not contribute). $\Box$

Theorem (Stein Interpolation) Let ${T_z}$ be a family of operators depending analytically on ${z}$ in the strip ${0 \leq \text{Re }z< 1}$ and continuous up to the boundary. Suppose also that

$\displaystyle \sup_{t \in {\mathbb R}} e^{-c_\alpha \cosh(\alpha t)} \Big(|| T_{0 + it}||_{L^{p_0} \rightarrow L^{q_0}} + ||T_{1 + it}||_{L^{p_1} \rightarrow L^{q_1}} \Big)< \infty$

for some ${c_\alpha > 0}$ and ${0 < \alpha < \pi}$. Then

$\displaystyle ||T_\theta||_{L^{p_\theta} \rightarrow L^{q_\theta}} < \infty$

where ${T_\theta}$, ${p_\theta}$, and ${q_\theta}$ were as in the statement of the Riesz–Thorin theorem.

Proof: Write ${\varphi}$ as above, note ${\log|\varphi|}$ is sub-harmonic. The result then follows from a careful revisiting of the 3-lines lemma; details to follow. $\Box$

Remark A “meta-application” of Stein is that one can show that if ${B: L^{p_0} \rightarrow L^{q_0}}$, ${ABC : L^{p_1} \rightarrow L^{q_1}}$, ${c^{it} : L^{p_1} \rightarrow L^{p_1}}$, ${A^{it} : L^{q_1} \rightarrow L^{q_1}}$, then ${A^\theta B C^\theta : L^{p_\theta} \rightarrow L^{q_\theta}}$. For example, ${A}$ (or ${C)}$ could be ${f \mapsto \cdot f}$ for fixed ${\varphi}$, say ${\varphi(x) = |x|^{\pm \alpha}}$.

Theorem If ${f \mapsto \hat f}$ maps continuously from ${L^p}$ to ${L^q}$, then ${q = p'}$ and ${1 \leq p \leq 2}$ (i.e., Hausdorff–Young is all the ${L^p}$-type estimates).

Proof: (a) Scaling. Let ${f \in \mathcal{S}({\mathbb R}^d)}$ be non-zero and fixed. Set ${f_\lambda = f(\lambda x)}$. Now ${||f_\lambda||_p = \lambda^{-\frac{d}{p}} ||f||_p}$. Also, ${\widehat{(f_\lambda)}(\xi) = \lambda^{-d} \hat f \left( \frac{\xi}{\lambda}\right)}$ so ${||\hat f_\lambda||_q = \lambda^{-d +\frac{d}{q}}||\hat f||_q}$. So if ${||\hat f||_{q} \lesssim ||f||_p}$, then we must have ${\lambda^{-d +\frac{d}{q}}||\hat f||_q \lesssim \lambda^{-\frac{d}{p}} ||f||_p}$. We can break this by sending ${\lambda \rightarrow 0}$ or ${\lambda \rightarrow \infty}$, unless ${\frac{1}{p} + \frac{1}{q} = 1}$. $\Box$

An interpretation: If ${f}$ is measured in ${[kg]}$ and ${x \sim [m]}$, then ${||f||_p \sim [kg][m]^\frac{d}{p}}$, ${\hat f \sim [kg][m]^d}$, ${\xi \sim [m]^{-1}}$, ${||\hat f||_q \sim [kg][m]^{d +\frac{d}{q}}}$.

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