# Integral kernels of self-adjoint operators [closed]

If the integral kernel $k(x, y)$ of an operator $T : C^\infty_c(M) \to \mathcal{D}'(M)$ is symmetric ($M$ is a compact manifold), then the operator $T$ is symmetric. Is the converse true? That is, given a self-adjoint $T$ which has an integral kernel $k(x, y)$, is $k(x, y) = k(y, x)$? A reference would be really appreciated.

## closed as off-topic by Christian Remling, Alex Degtyarev, coudy, Neil Strickland, Ryan BudneyMay 12 '15 at 20:56

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• It doesn't really make sense to talk about self-adjoint operators between two distinct spaces, but if you want to consider $T: L^2\to L^2$, then, yes, it's a standard fact that $T^*$ is an integral operator with kernel $k^*(x,y)=\overline{k(y,x)}$. See here: encyclopediaofmath.org/index.php/Integral_operator – Christian Remling May 12 '15 at 17:08

The following standard argument works if $k$ is assumed bounded. I don't think that assumption should be necessary, but maybe this is at least a helpful start.
If $T$ is symmetric then for every $f,g \in C^\infty(M)$ we have $$\int \int k(x,y) f(x) g(y)\,dx \,dy = \int \int k(x,y) g(x) f(y)\,dy\,dx$$ or changing variables and using Fubini's theorem (here we use the assumption that $k$ is bounded), $$\iint k(x,y) f(x) g(y) \,dx\,dy = \iint k(y,x) f(x) g(y)\,dx\,dy.$$ In other words, for every $F : M \times M \to \mathbb{R}$ which is of the form $F(x,y) = \sum_{i=1}^n f_i(x) g_i(y)$ where $f_i, g_i \in C^\infty(M)$, we have $$\iint (k(x,y) - k(y,x)) F(x,y) \,dx\,dy = 0.$$ Now using a monotone class argument, show that the same holds for all bounded measurable $F: M \times M \to \mathbb{R}$. Taking $F(x,y) = k(x,y) - k(y,x)$ you get $$\iint |k(x,y) - k(y,x)|^2 \,dx\,dy = 0.$$ (If $k$ is assumed continuous you can instead use Stone-Weierstrass in the last step.)