Show this function is differentiable at $0$ Let $f:\mathbb{R}\to \mathbb{R}$ be a compactly supported function that satisfies the Holder condition with exponent $\beta\in(0,1)$, i.e. we have a constant $A>0$ so that for all $x,y\in\mathbb{R}$, $|f(x)-f(y)|\le A|x-y|^{\beta}$. We want to show that $$g(x)=\int_{-\infty}^\infty \frac{f(y)}{|x-y|^\alpha} dy$$ is differentiable at $0$. Here $0<\alpha<\beta$.
 A: Just for the completeness of the problem, I will pose a solution myself.
So the idea is to use the so called generalized dominant convergence theorem, whose proof is exactly the same as the usual dominant convergence theorem. And the statement is the following:
Let $\{h_n\}$ be a family of $L^1$ function which converge almost everywhere to a $L^1$ function $h$, let $g_n$ be another family of $L^1$ functions such that $g_n\to g$ a.e., $|h_n|\le g_n$ for each $n$ and $|h|\le g$,  Suppose now we have $\lim \int g_n=\int g$, then $\lim\int h_n=\int h$.
Now come back to our problem. First attempt would be to use the standard test, which means we just naively take the derivative of the inside function and then check integrability. We immediately encounter a problem because we don't have the desired integrability condition. In fact, we have $f(y)\cdot sgn(x-y)|x-y|^{-\alpha-1}$, which is not integrable. 
So our second attempt would be to use the Holder condition to balance this singularity. Assume that $supp(f)\subset [-K,K]$. Let us first fix $x_n=\frac{1}{n}$, then for each $x_n=\frac{1}{n}$, there exists some $\xi_n$ where $0\le |\xi_n|\le |x|$ such that $\frac{1}{|y-x_n|^\alpha}-\frac{1}{|y|^\alpha}=\alpha\cdot sgn(y-\xi_n)\cdot |y-\xi_n|^{-\alpha-1}\cdot |x_n|$ by mean value property.
$$
\begin{aligned}
\frac{g(x)-g(0)}{x}&=\int_{-K}^K \frac{f(y)-f(\xi_n)}{|y-x|^\alpha\cdot x}-\frac{f(y)-f(\xi_n)}{|y|^\alpha\cdot x}+E_n\\
&=\int_{-K}^K \frac{f(y)-f(\xi_n)}{x}\cdot (\frac{1}{|y-x|^{\alpha}}-\frac{1}{|y|^\alpha})+E_n\\
&=\int_{-K}^K (f(y)-f(\xi_n))\cdot \alpha\cdot sgn(y-\xi_n)\cdot |y-\xi_n|^{-\alpha-1}+E_n
\end{aligned}
$$
where $E_n=\int_{-K}^K \frac{f(\xi_n)}{x}(\frac{1}{|y-x|^\alpha}-\frac{1}{|y|^\alpha})=(\int_{-K-x}^{-K}\frac{f(\xi_n)}{x}\frac{1}{|y|^\alpha}-\int_{K-x}^{K}\frac{f(\xi_n)}{x}\frac{1}{|y|^\alpha})$, this goes to zero as $x\to 0$ by mean value theorem.
It is now clear what the limit should be if we are able to pass the limit inside, it is exactly $\int_\mathbb{R} \chi\cdot(f(y)-f(0))\cdot \alpha\cdot sgn(y)\cdot |y|^{-\alpha-1}dy$ where $\chi$ is the characteristic function of $[-K,K]$. The integrand will be our function $h$.
We now justify this by applying the generalized dominant convergence theorem.
Fix $x=\frac{1}{n}$, then the integrand becomes: $h_n=\chi\cdot (f(y)-f(\xi_n))\cdot \alpha\cdot sgn(y-\xi_n)\cdot |y-\xi_n|^{-\alpha-1}$ where $0\le|\xi_n|\le |x|$ is a number which we get from mean value theorem. 
Now we define $g_n=C\cdot |y-\xi_n|^{\beta-\alpha-1}\cdot \chi$ where $C$ is some large constant. Similarly, we define $g=C\cdot |y|^{\beta-\alpha-1}\cdot \chi$. One easily see that $|h_n|\le g_n$, $|h|\le g$ by applying the Holder condition. Moreover, each $g_n$ is in $L^1$. Since $\xi_n\to 0$, we have $\lim \int g_n\to \int g$ and $g_n\to g$ a.e. Finally, invoke our generalized dominant convergence theorem, the problem is solved.
