# Show that $T$ is self adjoint where $Tf(x) := \int k(x,y)f(y)\,dy$ when $k(y,x) = \overline{k(x,y)}$.

The operator $Tf(x) := \int k(x,y)f(y)\,dy$ defined on $L^2(\mathbb{R}),$ where $k \in L^2 (\mathbb{R}^2))$ is self-adjoint when $k(y,x) = \overline{k(x,y)}$.

We need to show that for any $f,g \in L^2(\mathbb{R})$ that $$\langle Tf(x) , g(x) \rangle = \langle f(x), Tg(x) \rangle$$ We have that $$\langle Tf(x), g(x) \rangle = \int k(x,y)f(y)\,dy \int \overline{g(x)}\,dx = \iint k(x,y)f(y)g(x)\, dy\,dx$$ and $$\langle f(x), Tg(x) \rangle = \int f(x)\,dx \int \overline{k(x,y)g(y)} \,dy = \iint f(x) \overline{k(x,y)} \overline{g(y)} \,dx\,dy.$$ I have no idea how to show that these two integrals are equivalent with the additional condition on the function $k$. I imagine that I'm going about it the wrong way.

• You are missing the integral over $x$ in both cases. – Paul K Apr 28 '17 at 20:25
• Why isn't it $\overline{g(x)}$ in $\langle Tf(x),g(x)\rangle$? – Semiclassical Apr 28 '17 at 20:43
• I edited in between your comments. The answer below formulates it properly though... – Dragonite Apr 28 '17 at 20:48

Let's write things out using parentheses when necessary. Since $Tf(x)=\int k(x,y)f(y)\ dy$, we have $$\langle Tf,g\rangle =\int\left(\int k(x,y)f(y)\ dy\right)\overline{g(x)}\ dx$$ and $$\langle f, Tg\rangle=\int f(x)\overline{\left(\int k(x,y)g(y)dy\right)}\ dx.$$ To show that these are equal, start with $\langle Tf,g\rangle$, toy around with the expression, apply Fubini, use $k(x,y)=\overline{k(y,x)}$, toy around some more, and you're done.
• We can look at $T_a f(x) =1_{|x| < a}\int_{-a}^a k(x,y)f(y)dy$ and say it converges to $T$ when $a \to \infty$ – reuns Apr 28 '17 at 20:47