# Compact operator

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 Hilbert-Schmidt integral operator, can we conclude that $$K$$ is a compact operator. If yes, why?

• 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 Hilbert-Schmidt operator. – Zorngo Mar 3 at 23:49
• Can you please explain why does it converge if $k\in L^2([0,1]^2)$ ? – Samovem Mar 3 at 23:51
• @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 $L^2$? – Aleksei Kulikov Mar 4 at 0:09
• If you interchange the integrals, it becomes clear that this is exactly a Hilbert-Schmidt 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 $L^2([0,1]^2)$. – Nate Eldredge Mar 4 at 1:48