aplication of convergence theorem for integrals I am studying the proof of a theorem, and if my affirmation is true then I understand the proof of the theorem. I dont know  how to prove the affirmation...
Affirmation:
Let $n \in N , n \geq 1$ and let $0< \alpha < n$. Let $\varphi$ a smooth function in $R^n$ with compact support and $\int_{R^n} \varphi  = 1$ .Define $\varphi_{\epsilon }(y) = {\epsilon}^{-n}\varphi(y/ \epsilon)$
Then for a fixed $x \in R^n$
$$ \displaystyle\lim_{\epsilon \rightarrow 0^{+}} \displaystyle\int_{R^n}\frac{\varphi_{\epsilon }(y)}{|x-y|^{n-\alpha}}dy =  \frac{1}{|x|^{n-\alpha}}$$
I am trying to use the monotone converge theorem, the  dominated convergence theorem and that $\varphi $ is in the Schwartz space... This things its not working ..
Someone can give me a help to prove or disprove the affirmation?
thanks in advance
 A: Let's begin with the parameter transformation $y = \varepsilon z$. That gives us
$$\int_{\mathbb{R}^n} \frac{\varphi_\varepsilon(y)}{\lVert x-y\rVert^{n-\alpha}}\,dy = \int_{\mathbb{R}^n} \frac{\varphi(z)}{\lVert x-\varepsilon z\rVert^{n-\alpha}}\,dz.$$
Now there are two cases to distinguish. First, $x = 0$. Then we have
$$\varepsilon^{\alpha-n}\underbrace{\int_{\mathbb{R}^n} \frac{\varphi(z)}{\lVert z\rVert^{n-\alpha}}\,dz}_C,$$
and the limit depends on the value $C$, it is $+\infty$ if $C > 0$, $-\infty$ if $C < 0$, and $0$ if $C = 0$. If $\varphi \geqslant 0$, then we must have $C > 0$, but since that is not assumed, the other possibilities cannot be ruled out. So $x = 0$ is a point where the stated limit may not hold.
Next, consider $x\neq 0$. Then for small enough $\varepsilon > 0$ we have $x \neq \varepsilon z$ for all $z\in \operatorname{supp} \varphi$, and thus
$$\begin{align}
\left\lvert \int_{\mathbb{R}^n} \frac{\varphi(z)}{\lVert x-\varepsilon z\rVert^{n-\alpha}}\,dz - \frac{1}{\lVert x\rVert^{n-\alpha}}\right\rvert
&= \left\lvert\int_{\mathbb{R}^n} \varphi(z)\left(\frac{1}{\lVert x-\varepsilon z\rVert^{n-\alpha}} - \frac{1}{\lVert x\rVert^{n-\alpha}}\right)\,dz \right\rvert\\
&\leqslant \int_{\mathbb{R}^n} \lvert\varphi(z)\rvert\cdot\left\lvert\frac{1}{\lVert x-\varepsilon z\rVert^{n-\alpha}} - \frac{1}{\lVert x\rVert^{n-\alpha}}\right\rvert\,dz,
\end{align}$$
where the latter tends to $0$ by the dominated convergence theorem. If $\operatorname{supp}\varphi \subset B_R(0)$, and $\varepsilon < \frac{\lVert x\rVert}{2R}$, then
$$\left\lvert\frac{1}{\lVert x-\varepsilon z\rVert^{n-\alpha}} - \frac{1}{\lVert x\rVert^{n-\alpha}}\right\rvert \leqslant \frac{2^{n-\alpha}+1}{\lVert x\rVert^{n-\alpha}}.$$
So for $x\neq 0$, we have
$$\lim_{\varepsilon\searrow 0} \int_{\mathbb{R}^n} \frac{\varphi_\varepsilon(y)}{\lVert x-y\rVert^{n-\alpha}}\,dy = \frac{1}{\lVert x\rVert^{n-\alpha}},$$
and a more careful look reveals that the convergence is even locally uniform.
