A question from Stein's book, Singular Integral.

A question from Stein's book, Singular Integral. Let $\left\{ f_{m}\right\}$ be a sequence of integrable function such that $$\int_{% \mathbb{R}^{d}}\left\vert f_{m}\left( y\right) \right\vert dy=1$$ and its support converge to the origin: $$\text{supp} \left( f\right) =cl\left\{ x:f_{m}\left( x\right) \neq 0\right\}$$ In this text, by simple limiting argument we get $$\displaystyle \lim_{m\rightarrow \infty }\int_{\mathbb{R}^{d}}\frac{f_{m}\left( y\right) }{\left\vert x-y\right\vert ^{d-\alpha }}dy= \frac{1}{\left\vert x\right\vert ^{d-\alpha }}.$$

Could you explain me how to get the above result?

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Should there be $|f_m(y)|$ in the integral under limit? Could you give page number, just in case? –  user53153 Dec 27 '12 at 23:46
Could you fix Latex? Also what is the page number? –  Bombyx mori Dec 27 '12 at 23:56
@user32240: Not from exercise, but from page 119 –  beginner Dec 28 '12 at 0:07
@PavelM: Page 119 –  beginner Dec 28 '12 at 0:07
@user32240 I don't find anything horribly wrong with the latex, and I do find the relevant passages on page 119. Are you looking at the right book? It's Singular integrals and Differentiability properties of functions. To answer my own question: $f_m\ge 0$, so absolute values are not needed. –  user53153 Dec 28 '12 at 0:31

Some context: on page 119 Stein shows that the Riesz potential $I_\alpha$ is not bounded from $L^1$ to $L^{n/(n-\alpha)}$. The step in question is how to show that $(I_\alpha f_m)(x)\to |x|^{-n+\alpha}$ pointwise in $\mathbb R^n\setminus \{0\}$, where $\{f_m\}$ is a sequence that approximates the point mass at the origin. (Then, pointwise convergence + Fatou's lemma imply that $\liminf\|I_\alpha f_m\|_{n/(n-\alpha)}\ge \||x|^{-n+\alpha}\|_{n/(n-\alpha)}=\infty$, as claimed.)

Fix $x\in\mathbb R^n\setminus\{0\}$. By assumption the support of $f_m$ is contained in a small neighborhood of the origin, say $B_r=\{y: |y|<r\}$. For every $y\in B_r$ we have $|x-y|\le |x|+r$ by the triangle inequality. Hence $|x-y|^{-n+\alpha} \ge (|x|+r)^{-n+\alpha}$. Integration yields $$\int |x-y|^{-n+\alpha}f_m(y)\,dy \ge (|x|+r)^{-n+\alpha} \int f_m(y)\,dy=(|x|+r)^{-n+\alpha}$$. Since $r\to 0$ as $m\to \infty$, this already gives $\liminf_{m\to\infty} I_\alpha f_m(x)\ge |x|^{-n+\alpha}$, which is enough for Fatou. But just for sport, you can also get a bound from the other side and conclude that $\lim_{m\to\infty} I_\alpha f_m(x)=|x|^{-n+\alpha}$.

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What is the hint for get a bound from the other side? –  beginner Dec 29 '12 at 4:49
@beginner What we have so far was obtained from the triangle inequality $|x-y|\le |x|+r$. To get a reverse estimate, we should use... –  user53153 Dec 29 '12 at 4:57
Is it true that we use Holder Inequality? –  beginner Dec 29 '12 at 5:04
@beginner I was hinting at the reverse triangle inequality: $|x-y|\ge |x|-r$, and follow the steps as before with inequalities pointing the other way. You need $r<|x|$ for everything to go smoothly, which is not a problem because $x$ is fixed and $r$ goes to zero as $m$ grows. –  user53153 Dec 29 '12 at 5:06
so we get $lim sup I_\alpha f_m(x) \leq |x|^{-n+\alpha}$, thus we get the hypothesis of fatou. Thanks for the hint –  beginner Dec 29 '12 at 5:11

By a simple limit argument we assume that the simple $f_m$ is simply smooth and satisfies:

• $$\int_{\mathbf R^d} f_m = 1;$$
• $$\lim_{m \to \infty} f_m(x) = \delta(x).$$

So your simple equality states that $$\lim_{m \to \infty} \int_{\mathbf R^d} \frac{f_m(y)}{|x - y|^{d - \alpha}} \textrm{d}y = \int_{\mathbf R^d} \frac{\delta(y)}{|x - y|^{d - \alpha}} \textrm{d}y = \frac{1}{|x|^{d - \alpha}}.$$

Beware: Distributions need nice things to test against. So regularize!

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Yes, this problem is really a problem in distributions. –  Bombyx mori Dec 28 '12 at 0:38