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}


\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,


  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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