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Consider the following operator $\mathcal{L}: L^\infty(\mathbb{R})\rightarrow L^\infty(\mathbb{R})$: $$ \mathcal{L}v\equiv \int_{-\infty}^{\infty} K(x,y)v(y)dy. $$ Is this a compact operator if $K\in L^1(\mathbb{R}^2)\cap L^2(\mathbb{R}^2)\cap L^\infty(\mathbb{R}^2)$? My particular $K$ is bounded and decays exponentially fast for large $x$ and $y$.

If not, what kind of $K$ would make it compact?

If the answer is yes, does that mean it is compact on $L^\infty(\mathbb{R}^2)\cap L^2(\mathbb{R}^2)$, which is really the space I am interested in right now.

Frankly, on the one hand, I am not sure how to tackle that. On the other hand, this must be a well-known result, which I cannot find. I found this question, which almost answers: Integral operator on $L^p$ is compact

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3 Answers3

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In $L^1$ and $L^\infty$, information about the magnitude of the kernel (the exponential decay stated in the question) isn't enough to establish compactness: some kind of uniformity is also needed. For the operator described:

Assume that there is a constant $M$ such that for almost all $x\in\mathbb{R}$, $K(x,\cdot)\in L^1(\mathbb{R})$ and $\|K(x,\cdot)\|_1\leq M$ (that's a sufficient condition for boundeness of the operator on $L^\infty(\mathbb{R})$). Then the operator is compact iff for any $\varepsilon>0$ there exist $\delta>0$ and $R>0$ such that for a.a. $x\in\mathbb{R}$ and all $h\in(-\delta,\delta)$ we have $$\int_{\mathbb{R}\setminus(-R,R)}|K(x,y)|\mathrm{d}y<\varepsilon\tag{1}$$ and $$\int_{\mathbb{R}}|K(x,y+h)-K(x,y)|\mathrm{d}y<\varepsilon\tag{2}$$ These two conditions on cross-sections through the kernel require (1) uniform decay at $\pm\infty$ and (2) uniform continuity of translation.

See my paper in Proc. Amer. Math. Soc. 123 (1995), 3709-3716 for details.

On $L^\infty\cap L^2$, presumably the intended norm is the sum, maximum or similar of the $L^2$ and $L^\infty$ norms? If so, a straightforward subsequence argument shows that any kernel that gives rise to a compact operator on $L^2$ and a compact operator on $L^\infty$ will give rise to a compact operator on the intersection. The Hilbert-Schmidt condition should be enough for $L^2$ compactness in this instance.

Simon Eveson
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If $K \in L^2(\mathbb{R}^2)$ then this is a compact hilbert-schimdt operator, as stated in wikipedia: Hilbert–Schmidt integral operator.
You can show it is compact if you approximate $K$ with polynomials.

GuySa
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I found this paper, which gives the necessary and sufficient condition on $K$ for the operator to be compact in Corollary 5.1 at the end:

S. P. Eveson, Compactness Criteria for Integral Operators in $L^\infty$ and $L^1$ Spaces, Proceedings of the American Mathematical Society 123, 1995, 3709-3716.

https://www.ams.org/journals/proc/1995-123-12/S0002-9939-1995-1291766-8/S0002-9939-1995-1291766-8.pdf

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