The statement is like this,
$K\subset\mathbb{R}$ is compact, the operator $A:L^\infty(K)\mapsto L^\infty(K)$ is defined by $f(x)\mapsto\int_K k(x,y)f(y)dy$. For $x\neq y$, $|k(x,y)|\leq M|x-y|^{\alpha-2}$, where $\alpha\in(0,2]$ and $M$ is constant. Show $A$ is a compact operator.
I thought about Hilbert-Schmidt operator first. However, I noted that, although we can relax $L^\infty(K)$ to $L^2(K)$ (because $L^\infty\subset L^2$, is that correct?), $k\notin L^2(K\times K)$. So the condition for Hilbert-Schmidt is not satisfied.
Information about $k$ is not enough, so it seems difficult to construct $\{k_n\}$ that makes $A_n$ compact and converge to $A$ – so that $A$ is compact. Therefore, now I am thinking about Arzela-Ascoli.
So, I think I should take a bounded sequence $f_n(x)$ in $L^\infty(K)$, and prove that $Af_n$ is uniformly bounded and equicontinuous.
For uniformly boundedness,
$$
||Af_n||_\infty=\sup_{x\in K}|\int_Kk(x,y)f_n(y)dy|\leq\sup_{x\in K}\int_KM|x-y|^{\alpha-2}|f_n(y)|dy
$$
Then I get stuck in how to obtain the bound.
For equicontinuity,
$$
||Af_n(x_1)-Af_n(x_2)||_\infty=\sup_{x_1,x_2\in K}|\int_K[k(x_1,y)-k(x_2,y)]f_n(y)dy|\leq\sup_{x_1,x_2\in K}\int_K|k(x_1,y)-k(x_2,y)||f_n(y)|dy
$$
Again I got stuck in bounding the integral.
So, is my general idea correct? If so, how should I proceed to bound the two integrals? Any helpful suggestion will be appreciated!
Thanks in advance!
Update:
I got an idea in using sequence of compact operators. Maybe we can consider $k_n(x,y)$ such that $|k_n|\leq \frac{M}{|x-y|^{2-\alpha}+\delta}$, where $\delta>0$. $A_n$ can be defined by using $k_n$ as kernel. It seems $k_n\in L^2(K\times K)$ and $A_n$ is Hilbert-Schmidt integral operator, then $A_n$ is compact. Then we only need to prove that $A_n\rightarrow A$.
I am working on details of that idea.
By the way, the bound for $k(x,y)$ in the statement is correct. Actually, that form is common in Green's functions.
Best Answer
If all you know about $f$ is that it is in $L^\infty(K)$, the best possible bound you can have on integrals of $f$ times some other function is that $$ \left|\int_K g(x) f(x)\; dx \right| \le \|f\|_\infty \int_K |g(x)|\; dx$$
There's a problem with your given $k(x,y)$, though: $|x - y|^{\alpha - 2}$ has a non-integrable singularity at $x=y$ if $\alpha \le 1$, i.e. in that case $$\int_{x-\epsilon}^{x+\epsilon} |x - y|^{\alpha - 2}\; dy = \infty$$ Are you sure you copied the problem correctly? It's not $\alpha - 1$ rather than $\alpha - 2$?