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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

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