# 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!

• Your only hypothesis is that $Tf(y)$ is well-defined for a.e. $y\in\mathbb R$ whenever $f\in L^2$ and the operator $T$ is bounded on $L^2$? Or do you have additional assumptions on the kernel $k$? Commented May 17, 2014 at 18:06
• @Etienne Only that it is well-defined pointwise almost everywhere on $L^2(\mathbb{R})$ and that the operator is bounded. The kernel is, of course, measurable though. For instance, it is not assumed that the kernel is Hilbert-Schmidt. Commented May 17, 2014 at 18:07
• OK, thanks. One natural idea would be to show that interchanging the integral is justified at least say for bounded functions with compact support (and the conclude by a "density" argument); but under the "minimal" assumption, this does not seem completely obvious. Perhaps it is... Commented May 17, 2014 at 18:10
• @Etienne Yeah I had considered that but I was really hoping to avoid that kind of argument. I was hoping for something a bit more high-powered. Commented May 17, 2014 at 18:14
• One more question about "well-defined": is the integral defining $Tf(y)$ assumed to be "absolutely convergent" (for a.e. $y$), or can it be even trickier (with some kind of principal value or whatever)? Commented May 17, 2014 at 18:17

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

$$\int_{\mathbb R}\int_{\mathbb R} \vert k(x,y)\vert \, \vert f(x)\vert \,\vert g(y)\vert \, dxdy=\langle T_{\vert k\vert}\vert f\vert,\vert g\vert\rangle_{L^2}\, .$$

• This was along the lines I was thinking as well. The more general case still eludes me but it could be some minor alteration of this argument. Commented May 21, 2014 at 11:52
• @Cameron I really don't know. Anyway, I posted a question about this: math.stackexchange.com/questions/800570/… Commented May 21, 2014 at 18:45
• I've found a proof (more or less) in Halmos and Sunder. It's Theorem 7.5. Commented May 27, 2014 at 4:14
• @CameronWilliams That's great! Commented May 27, 2014 at 4:55
• The theorem is actually incredibly inobvious. I would not have thought about doing it that way. If you ever feel so inclined, you can check it out. It relies heavily on the immediately preceding theorem. Commented May 27, 2014 at 5:16