Why Does Fubini-Tonelli Apply Here?

Question

Let $H: [0,1]^2 \rightarrow \mathbb{R}$ be symmetric and measurable. Suppose that, for every $f \in L^2(0,1)$

• $H(x, \cdot) f : y \mapsto H(x, y) f(y)$ is integrable for almost all x

• the function $Tf$ given by $$(Tf)(x) = \int_0^1 H(x, y) f(y) dy, \qquad x \in (0,1)$$ belongs to $L^2(0,1)$

We find (in an earlier part of the question) that $T$ is a bounded linear operator and want to show that it is self-adjoint. However, a key step in showing “self-adjointness” is the following step:

For any $g \in L^2(0,1)$, $$ \begin{align} & \int_0^1 \left ( \int_0^1 H(x,y)f(y) \, dy \right ) \overline{g(x)} \, dx \\ = \, &\int_0^1 \left ( \int_0^1 H(x,y)\overline{g(x)} \, dx \right ) f(y) \, dy \end{align} $$

However, I can’t find a reason to justify the change in the order of integration here.

Answer 1

I think it all boils down to showing that $H\in L_1((0,1)\times(0,1))$.

This is just a suggestion. I have not checked the details and it may still fail:

1. Define $\mathcal{E}$ the set of functions of the form $\sum^n_{k=1}\phi_k(x)\psi_k(y)$ where $\phi_k$ and $\psi_k$ are simple functions. On $\mathcal{E}$ define the linear functional $$I(\Psi)=\int^1_0\Big(\int^1_0H(x,y)\Psi(x,y)\,dy\Big)\,dx$$
2. Try to show that $I$ is dominated by some constant multiple of the in $L_2((0,1)\times(0,1))$, that is $$|I(\Psi)|\leq c\|\Psi\|_2$$
3. If (2) works, then one canextend $I$ to all of $L_2$ by taking limits in of functions in $\mathcal{E}$ in $L_2((0,1)\times(0,1))$. That this can be done needs to ve verified.

In particular we can take $\Psi(x,y)=\operatorname{sign}(H(x,y))$. Fatou's lemma will then imply that $|H(x,y)|\in L_1((0,1)\times(0,1))$.