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