Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 down vote accepted

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 $\lim_{m\to\infty} I_\alpha f_m(x)=|x|^{-n+\alpha}$.

share|cite|improve this answer
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 $$\lim_{m \to \infty} \int_{\mathbf R^d} \frac{f_m(y)}{|x - y|^{d - \alpha}} \textrm{d}y = \int_{\mathbf R^d} \frac{\delta(y)}{|x - y|^{d - \alpha}} \textrm{d}y = \frac{1}{|x|^{d - \alpha}}.$$

Keywords: Friedrichs mollifier; mollifier; approximative identity; approximation to the identity; nascent delta function

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

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.