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February 4, 2011

Harmonic Analysis Lecture Notes 11

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:55 pm

Translation Invariant Operators {\equiv} Fourier Multipliers {\equiv} Convolution Operators

Theorem (Hardy–Littlewood–Sobolev inequality, Fractional Integration, Weak Young) Let {g \in L^{q,\infty}({\mathbb R}^d)} and {1 < p < r < \infty} and {\frac{d}{r} + d = \frac{d}{q} + \frac{d}{p}}. Then

\displaystyle  	||g * f||_{L^r} \lesssim_{d,p,q} ||f||_{L^p} ||g||_{L^{q,\infty}}^*

In particular,

\displaystyle  	\Big|\Big| ||x|^{\alpha - d} * f \Big|\Big|_{L^r} \lesssim || f||_{L^p}

for {0 < \alpha< d} and {\frac{d}{r} + \alpha = \frac{d}{p} + \frac{d}{q}}.

Remark

  1. Convolution (generically) makes functions shorter and fatter. Consequently, if a convolution operator maps {L^p \rightarrow L^r}, then {r \geq p}.
  2. If {\alpha \leq 0}, then {|x|^{\alpha -d}} is not even a distribution!

Personal Comment: The Hedberg proof of this result was presented in a participating analysis seminar here at UCLA, and thus was not given in lecture. The transcription of that proof will be given shortly.

Proof: By Marcinkiewicz, if we prove {||f * g||_{L^{r,\infty}} \lesssim ||g||_{L^{q,\infty}} ||f||_{L^p}} then the result will still follow (we are necessarily dealing with an open set of exponents). (Indeed this weak-type statement is even true when {p=1}-this is the {\Delta}-inequality in {L^{r,\infty}}.) Given {\lambda} we split {g} into

\displaystyle  	g = g_1 + g_2 = g\cdot \chi_{|g|> \mu} + g \cdot \chi_{|g| \leq u}

the relation between {\mu} and {\lambda} will be given later.

Now, {||f * g_2||_\infty \leq ||f||_{L^p} ||g_2||_{L^{p'}}} and

\displaystyle  \begin{array}{rcl}  	||g_2||_{L^{p'}}^{p'} &=& \displaystyle \int_0^\mu p' h^{p'-1} \underbrace{|\{ |g_2| > h\}|}_{h^{-q} ||g||^q_{L^{q,\infty}}} \; dh \\ \\ 	\scriptsize\begin{bmatrix} 	\text{needed} q < p' \\ 	\text{or} \frac{d}{p'} = d - \frac{d}{p} < \frac{d}{q} 	\end{bmatrix}		&=& \displaystyle ||g||^{q}_{L^{q,\infty}} \cdot \frac{p'}{p'-q} \mu^{p'-q} 	\end{array}

Therefore

\displaystyle  	||f* g||_{L^\infty} \lesssim ||f||_{L^p} ||g||^{q/p'}_{L^{q,\infty}} \mu^\frac{p' - q}{p'}.

Recall that we are trying to show {|\{ (f*g) > \lambda\}| \lesssim \lambda^{-r} ||f||_{L^p}^r \big( ||g||_{L^{p,\infty}}^*\big)^r}. But

\displaystyle  	|\{ |f * g| > \lambda \}| \leq \big| \{ |f * g_1| > \frac{\lambda}{2}\}\big| + \big| \{ |f * g_2| > \frac{\lambda}{2}\}\big|

From the computation above, if we choose {\mu} wisely we can make the second set empty, say

\displaystyle  	\mu = C \lambda^{r/q}

for some constant {C}. We’re stuck trying to show

\displaystyle  	|\{ |f*g_1|> \frac{\lambda}{2} \} | \lesssim \lambda^{-r} ||f||_{L^p}^r ||g||_{L^{q,\infty}}^r.

Well,

\displaystyle  	LHS \lesssim \frac{||f *g_1||_{L^p}^p}{\lambda^p} \leq \lambda^{-p} ||f||_{L^p}^p ||g_1||_{L^1}^p

To conclude we note

\displaystyle  \begin{array}{rcl}  	||g_1||_{L^1} &=& \displaystyle \int_\mu^\infty |\{|g_1| > h \}| \; dh \\ \\ 		&\lesssim& \displaystyle ||g||^q_{L^{q,\infty}} \int_\mu^\infty h^{-q} \; dh \\ \\ 		&\lesssim& ||g||^q_{L^{q,\infty}} \mu^{1-q} 	\end{array}

As {\mu^{p(1-q)} \propto \lambda^{rp(1-q)}{q} = \cdots = \lambda^{p-r}}, we are done. \Box

What is the Fourier multiplier associated to {(|x|^{\alpha -d} * )}? I.e., what is the (distributional) Fourier transform of {|x|^{\alpha - d}}? Note that it must be spherically symmetric, and homogeneous of degree {-\alpha}.

\displaystyle \begin{matrix} 	\displaystyle e^{-\pi|x|^2} &\stackrel{FT}{\longrightarrow}& \displaystyle e^{-\pi|\xi|^2} \\ 	\displaystyle e^{-\pi t|x|^2} &\stackrel{F}{\longrightarrow}& \displaystyle t^{-d/2} e^{-\pi|\xi|^2/t} \\ \\ 	\displaystyle \int_{\epsilon}^{1/\epsilon} e^{-\pi t|x|^2} t^\frac{d-\alpha}{2} \frac{dt}{t} &\stackrel{FT}{\longrightarrow}& \displaystyle \int_\epsilon^{1-\epsilon} e^{-\pi|\xi|^2/t} t^{-\frac{\alpha + d}{2}} \frac{dt}{t} 	\end{matrix}

The left hand side of the last expression converges, distributionally, as {\epsilon \downarrow 0} to

\displaystyle  	(\pi|x|^2)^\frac{\alpha -d}{2} \int_0^\infty e^{-u} u^\frac{d-\alpha}{2} \; \frac{du}{u} = \pi^{-\frac{d-\alpha}{2}} \Gamma\left(\frac{d-\alpha}{2}\right) |x|^{\alpha -d}

and the right hand side converges in the same manner to

\displaystyle  	(\pi |\xi|^2)^{-\frac{\alpha}{2}} \int_0^\infty e^{-u} u^{-\frac{\alpha}{2}} \; \frac{du}{u} = \pi^{-\frac{\alpha}{2}} \Gamma\left(\frac{\alpha}{2} \right) |\xi|^{-\alpha}.

Thus we indeed have

\displaystyle  	|x|^{\alpha-d} \stackrel{FT}{\longrightarrow} c_\alpha |\xi|^{-\alpha}

where

\displaystyle  	c_\alpha = \frac{\pi^{-\frac{\pi}{2} + \frac{d-\alpha}{2}} \Gamma\left(\frac{\alpha}{2}\right)}{\Gamma\left(\frac{d-\alpha}{2}\right)}.

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