# How to show a specific operator is compact?

Let $\Omega$ be a bounded region in $\mathbb{R}^N$ with $N\geq 1$.

Let $C(\overline{\Omega})$ denote the collection of all real-valued continuous functions $f \colon \overline{\Omega} \to \mathbb{R}$.

Define an operator $M \colon C(\overline{\Omega}) \to C(\overline{\Omega})$ by $$(Mv)(x) := \int_{\Omega} \left[ v(y)\right] k(x,y) \mathrm{d}y , \qquad (x \in \overline{\Omega}).$$

Suppose that the kernel $k \colon \overline{\Omega} \times \overline{\Omega} \to \mathbb{R}$ is continuous and nonnegative.

My question is that under what additional conditions, the operator $M$ will be compact?

Any idea or suggestions are much appreciated! Thank you:)

• What do you mean by compact for a non linear operator? – copper.hat Dec 15 '17 at 5:52
• Thanks @copper.hat. Let $X$ and $Y$ be normed spaces over $\mathbb{R}$. The operator $A \colon Z \subset X \to Y$ is called compact if and only if (1) $A$ is continuous; and (2) $A$ transforms bounded sets into relatively compact sets. – Paradiesvogel Dec 15 '17 at 5:56
• In other words, condition (2) is equivalent to saying that: if $( v_n)$ is a bounded sequence in $Z$, then there exists a subsequence $(v_{n_k})$ of $( v_n)$ such that the sequence $(A v_{n_k})$ is convergent in $Y$. – Paradiesvogel Dec 15 '17 at 6:01
• What norm are you using on $C(\overline{\Omega})$? – copper.hat Dec 15 '17 at 15:13
• Thanks @copper.hat. The norm should be supremum norm. – Paradiesvogel Dec 15 '17 at 19:39

I think this operator is already compact on $C(\overline{\Omega})$.
Let $A$ be a bounded subset of $C(\overline{\Omega})$; in particular we may assume $A$ is contained in the ball of radius $R$ in $C(\overline{\Omega})$. We would like conditions on $\phi$ so that $M$ sends $A$ to a precompact set. The typical way to accomplish this is to employ Ascoli-Arzela: i.e. prove equicontinuity and pointwise boundedness of the set $M(A)$. Let's see where this goes.
For pointwise boundedness, let $x\in\overline{\Omega}$. We have $$|(Kv)(x)| \leq \int_\Omega |\phi[v(y)]||k(x,y)|dy \leq \|\phi\|_{L^\infty[-R,R]}\|k\|_{L^\infty(\Omega\times\Omega)}|\Omega| < \infty,$$ where $|\Omega|$ denotes the Lebesgue measure of $\Omega$. Here we have used the continuity of $\phi$ and $k$. Since this bound is independent of $x$, applying $\phi^{-1}$ and using the continuity of $\phi^{-1}$ then leads to bounds on $|(Mv)(x)|$ independent of $x$, that is $M(A)$ is pointwise bounded in $C(\overline{\Omega})$.
For equicontinuity, let $\varepsilon>0$. Uniform continuity of $k$ on $\overline{\Omega}\times\overline{\Omega}$ implies implies that for any $\varepsilon'>0$ and $x,y\in\overline{\Omega}$ with $|x-y|$ sufficiently small, $$|(Kv)(x) - (Kv)(y)| \leq \int_\Omega |\phi[v(z)]||k(x,z) - k(y,z)| dz \leq \|\phi\|_{L^\infty[-R,R]}|\Omega|\varepsilon'.$$ Then provided we have uniform continuity of $\phi^{-1}$ on the set of all possible values of $(Kv)(x)$, taking $\varepsilon'$ sufficiently small we could show that $$|\phi^{-1}((Kv)(x)) - \phi^{-1}((Kv)(y))| < \varepsilon.$$ But the set of all possible values of $(Kv)(x)$ is contained in a compact interval, so $\phi^{-1}$ is automatically uniformly continuous on this set. These estimates in fact establish uniform equicontinuity.