# Compactness and boundedness of integral operator

I got some trouble with my homework question :

Let $B$ be the unit ball in $\mathbb{R}^d$, and let $T$ be an integral operatpor on $L^2(B)$ with kernel $K(x,y)$.

1. Suppose that $\sup_x \int_B |K(x,y)|\ dy \leq A$ and $\sup_y \int_B |K(x,y)|\ dx \leq A$ . Show that $\|T\|\leq A$.
Hint: Use the characterisation $$\|T\| = \sup_{\|f\|=\|g\|=1}|\langle Tf , g\rangle|$$ and use Cauchy-Schwarz and the given two conditions on the resulting double integrals.

2. Suppose that $K(x,y)=|x-y|^{-d+\alpha}$, where $x,y \in B$ and $\alpha>0$. Show that $T$ is a bounded operator on $L^2(B)$.

3. Show that under the same assumption as in 2) that $T$ is compact.

Hint: Consider the integral operator $T_n$ with kernel $K_n(x,y)= |x-y|^{-d+\alpha}\chi_{|x-y|>\frac 1n},$ where $\chi_{|x-y|>\frac{1}{n}}$ is the characteristic function of the set $\{|x-y|>1/n\}$.

For part (1), my working is \begin{align} ||Tf||^2&= \int_B |Tf(x)|\ ^2dx\\ &= \int_B |\int_BK(x,y)f(y)dy|\ ^2dx\\ &= \int_B |\int_B|K(x,y)|^ {1/2}|K(x,y)|^ {1/2}f(y)dy|\ ^2dx \\ &\leq \int_B (\int_B|K(x,y)|dy)(\int|K(x,y)||f(y)|^2dy)dx \mbox{ (by Cauchy-Schwarz}) \\&\leq \int_B A\int|K(x,y)||f(y)|^2dy)dx \\ &= A\int_B|f(y)|^2\int_BK(x,y)dy\\ &\leq A^2\int_B|f(y)|^2dy\\ &=A^2||f||^2. \end{align}

Hence, by $||Tf||^2$ ≤ $A^2||f||^2$ ，we have ||T||≤ A .

I am not sure if it is right , cus I didn't use the condition $\lVert T\rVert= \sup|(Tf , g)|$ where $\lVert f\rVert=1$ and $\lVert g\rVert= 1$.

For part (2) , I guess if I should show $||k||^2=\int_B \int_B|k(x,y)|dxdy$ is bounded by a finite $C$, then T is bounded by this $C$. However, I don't know how to show this double integral is bounded . I know the fact that in $R^d$, $\int_B1/|x|^{\alpha}$ is finite for $\alpha\leq d$，and escape to infinity otherwise. I guess this fact might be helpful for solving this part.

I don't know how to do (3) either. Guess I have to show $T_n$ converges to $T$ in the operator norm .

Can somebody help me to mend my working and show how to do (2) and (3) ? Thanks in advance.

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Hello, I have typeset equations in your question statements (1), (2), and (3). Please verify that I have gotten the statements correct. –  Neal Aug 11 '12 at 14:36
This is a homework problem from MATH3325 at the Australian National University. Online discussion of homework problems is allowed, but only with proper attribution (both when posting online, and when submitting work). –  Scott Morrison Aug 15 at 1:20

1. You're right, in fact, if you use Hölder's inequality instead of Cauchy-Schwarz you get that the operator is bounded in $L^p$ for every $1\leq p \leq \infty$.
2. The idea is to use 1 and for this, by symmetry, we only have to check that $$\sup_x \int_B |x-y|^{-d+\alpha}dy \leq A$$ for some $A>0$. We prove that $A= \int_B |y|^{-d+\alpha}dy$ works for $-d+\alpha<0$, the bound is easier otherwise since there is no singularity. (This is geometrically obvious, since the integral will be larger where the singularity is far away from the boundary, and the largest then at $x=0$) Notice that \begin{align} \int_B |x-y|^{-d+\alpha} dy = \int_{B_1(x)} |z|^{-d+\alpha} dz \end{align} and so \begin{align} \int_B |y|^{-d+\alpha} dy -\int_B |x-y|^{-d+\alpha} dy & = \int_{B\setminus B_1(x)} |y|^{-d+\alpha} dy - \int_{B_1(x)\setminus B} |y|^{-d+\alpha} dy \\ & \geq |B\setminus B_1(x)| - |B_1(x)\setminus B| =0 \end{align} where the last inequality follows since $|y|^{-d+\alpha} \geq 1$ on the first set and $\leq 1$ on the second.
3. Clearly the $K_n$ define compact operators (since $K_n \in L^2(B\times B)$), to prove the convergence in norm we again use 1. Define $L_n = K-K_n$ then \begin{align} \int_B |L_n(x,y)|dy & = \int_B |x-y|^{-d+\alpha}(1-\chi_{|x-y|>1/n}) dy = \int_B |x-y|^{-d+\alpha} \chi_{|x-y|\leq 1/n} dy\\ &\leq \int_B |y|^{-d+\alpha} \chi_{|y|\leq 1/n}dy =\int_{B_{1/n}} |y|^{-d+\alpha}dy. \end{align} where the last inequality is by the same argument as in 2 (this assumes $-d+\alpha<0$ the other case is easier since we already have $K\in L^2(B\times B)$). Now the last integral you can calculate explicitly to be $\frac{d|B|}{\alpha n^{\alpha}} \to 0$ as $n\to \infty$. By 1 this implies that $\|L_n\| \to 0$.
@lindamac: I've edited to add the proof of the bound. The last part is pretty straightforward, we proved that $$\sup_x\int_B |L_n(x,y)|dy \leq \frac{d|B|}{\alpha n^\alpha}$$ so by 1 we have that $\| L_n\| \leq \frac{d|B|}{\alpha n^\alpha}$ which implies that $\| K-K_n\| \to 0$ as $n\to \infty$. –  Jose27 Aug 12 '12 at 6:57