Is $L^2(0,\infty;L^2(\Omega)) = L^2((0,\infty)\times \Omega)$?

Question

If $\Omega$ is a bounded $C^1$ domain, is $L^2(0,\infty;L^2(\Omega)) = L^2((0,\infty)\times \Omega)$? Are they the same?

I know this is true when instead of $(0,\infty)$ we have a bounded interval.

Answer 1

Its not completely clear what you mean with "the same". You probably mean that the map

$$ \Phi : L^2 ((0,\infty) \times \Omega) \to L^2((0,\infty);L^2(\Omega)), f \mapsto (x \mapsto f(x,\cdot)) $$

is an isometric isomorphism.

That is indeed the case. To see this, let $f$ be a step function of the form $$f(x,y) = \sum_n \alpha_n \chi_{A_n} (x) \chi_{B_n}(y). \qquad (\dagger)$$

Since the Borel-$\sigma$-Algebra on $(0,\infty)\times\Omega$ is the product $\sigma$-Algebra, one can show that functions of this form are dense in $L^2((0,\infty)\times\Omega)$. In this case, a Fubini argument shows

$$ \Vert \Phi(f) \Vert^2 = \int_0^\infty \Vert f(x,\cdot)\Vert^2 \, dx = \int_0^\infty \int_\Omega |f(x,y)|^2 \, dx \, dy = \Vert f \Vert_2^2. $$

This means that $\Phi$ is well-defined on the space of step functions of the above form and thus extends uniquely to an isometric map $\tilde{\Phi} : L^2((0,\infty) \times \Omega) \to L^2((0,\infty); L^2(\Omega))$.

Since the set of step functions $g(x) = \sum_n f_n \chi_{A_n}(x)$ with $f_n = \sum_m a_m^{(n)} \chi_{B_m}$ forms a dense subset of $L^2((0,\infty);\Omega)$ (why?), we see that $\tilde{\Phi}$ forms an isometric isomorphism.

All we have to check is $\tilde{\Phi} = \Phi$. To see this, let $f \in L^2((0,\infty)\times \Omega)$ and choose a sequence $(f_n)_n$ of step functions as in $(\dagger)$ with $f_n \to f$ in $L^2((0,\infty) \times \Omega)$. By passing to a subsequence, we can assume $f_n (x,y) \to f(x,y)$ almost everywhere as well as $\tilde{\Phi}(f_n)(x) \to \tilde{\Phi}(f)(x)$ almost everywhere.

By a Fubini-argument, we see that for almost every $x$, we have $f_n(x, y) \to f(x,y)$ for almost all $y$. Fix some $x$ with this property and with $\tilde{\Phi}(f_n)(x) \to \tilde{\Phi}(f)(x)$ (this holds for almost all $x$).

By passing to a further subsequence (possibly depending on $x$!), we get $$f_{n_k}(x,y) = \tilde{\Phi}(f_{n_k})(x)(y) \to \tilde{\Phi}(f)(x)(y)$$ for almost all $y$. But the left hand side also converges to $f(x,y)$ for almost all $y$. Hence, $f(x,y) = \tilde{\Phi}(f)(x)(y)$ for almost all $y$ for almost all $x$.

This completes the proof.