Let $k:[0,1]^2 \to [0,1]$ be a measurable function. Define $K:L^2([0,1])\to L^2([0,1])$ to be the operator: $$ (Kf)(x) = \int_0^1\int_0^1 f(z) k(x,y) \mathbf{1}_{x\leq z\leq y} \ \mathrm{d}z \mathrm{d}y $$ $K$ seems like a HilbertSchmidt integral operator, can we conclude that $K$ is a compact operator. If yes, why?
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$\begingroup$ For a general measurable function $k : [0,1]^2 \to [0,1]$ the integral you have defined may not converge. If, on the other hand, $k$ belongs to $L^2([0,1]^2)$ then you get a HilbertSchmidt operator. $\endgroup$ – Zorngo Mar 3 at 23:49

$\begingroup$ Can you please explain why does it converge if $k\in L^2([0,1]^2)$ ? $\endgroup$ – Samovem Mar 3 at 23:51

2$\begingroup$ @Zorngo Umm, may be I'm missing something, but isn't any bounded function ($k:[0, 1]^2\to [0, 1]$) on the set of finite measure is in $L^2$? $\endgroup$ – Aleksei Kulikov Mar 4 at 0:09

3$\begingroup$ If you interchange the integrals, it becomes clear that this is exactly a HilbertSchmidt integral operator $Kf(x) = \int_0^1 f(x)\kappa(x,z)\,dz$ where $\kappa(x,z) = \int_0^1 k(x,y) 1_{x \le z \le y}\,dy = 1_{x \le z} \int_z^1 k(x,y)\,dy$ is bounded and in particular is in $L^2([0,1]^2)$. $\endgroup$ – Nate Eldredge Mar 4 at 1:48