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