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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!

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    $\begingroup$ 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 $k$? $\endgroup$
    – Etienne
    Commented May 17, 2014 at 18:06
  • $\begingroup$ @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. $\endgroup$ Commented May 17, 2014 at 18:07
  • $\begingroup$ 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... $\endgroup$
    – Etienne
    Commented May 17, 2014 at 18:10
  • $\begingroup$ @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. $\endgroup$ Commented May 17, 2014 at 18:14
  • $\begingroup$ 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)? $\endgroup$
    – Etienne
    Commented May 17, 2014 at 18:17

1 Answer 1

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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

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

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  • $\begingroup$ 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. $\endgroup$ Commented May 21, 2014 at 11:52
  • $\begingroup$ @Cameron I really don't know. Anyway, I posted a question about this: math.stackexchange.com/questions/800570/… $\endgroup$
    – Etienne
    Commented May 21, 2014 at 18:45
  • $\begingroup$ I've found a proof (more or less) in Halmos and Sunder. It's Theorem 7.5. $\endgroup$ Commented May 27, 2014 at 4:14
  • $\begingroup$ @CameronWilliams That's great! $\endgroup$
    – Etienne
    Commented May 27, 2014 at 4:55
  • $\begingroup$ 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. $\endgroup$ Commented May 27, 2014 at 5:16

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