Q:

Adjoint of an integral operator

I'm reading through a text about integral operators and I've come across the following theorem:

Let $k:\mathbb{R}^2\rightarrow\mathbb{C}$ be a kernel, $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ be a bounded operator given by $$Tf(y) = \int_{-\infty}^{\infty} k(x,y)f(x)\,dx.$$ Then the adjoint, $T^*$, of $T$ is given by $$T^*f(y) = \int_{-\infty}^{\infty}\overline{k(y,x)}f(x)\,dx.$$

The proof is as follows:

$$\begin{align} \langle Tf,g\rangle & \stackrel{\text{def}}{=} \langle f,T^*g\rangle \\ &= \int_{-\infty}^{\infty} Tf(y)\overline{g(y)}\,dy \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} k(x,y)f(x)\,dx\, \overline{g(y)}\,dy \tag{1} \\ &= \int_{-\infty}^{\infty} f(x) \int_{-\infty}^{\infty} k(x,y)\overline{g(y)}\,dy\,dx \tag{2} \\ &= \int_{-\infty}^{\infty} f(x)\overline{\int_{-\infty}^{\infty}\overline{k(x,y)}g(y)\,dy}\,dx \end{align}$$

From there, you can easily identify that the second piece in the integral would be the adjoint, by definition. My question is: how is the change of integral justified from $(1)$ to $(2)$? I'm sure it's some Fubini-type argument incorporating boundedness of the operator but I don't see how to make it precise.
Thanks!

A:

OK, here is an answer under a slightly stronger assumption.
Precisely, let's assume that the kernel $\vert k(x,y)\vert$ defines a bounded operator on $L^2$; and let us denote by $T_{\vert k\vert}$ this operator
To sow that the use of Fubini is justified, we have to check that 
$$ \int_{\mathbb R}\int_{\mathbb R} \vert k(x,y)\vert \, \vert f(x)\vert \,\vert g(y)\vert \, dxdy<\infty\, .$$
But this is ... trivial, since we have
$$\int_{\mathbb R}\int_{\mathbb R} \vert k(x,y)\vert \, \vert f(x)\vert \,\vert g(y)\vert \, dxdy=\langle T_{\vert k\vert}\vert f\vert,\vert g\vert\rangle_{L^2}\, .
$$

