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How can we identify $L^2(Q_T)$ and $L^2(0, T; L^2(\Omega))$?

For my understanding: $Q_T:=(0, T)\times \Omega$;

DEF1: $L^2(Q_T)=\{u: (0, T)\times \Omega \to \mathbb{R}, \mbox{measurable and} \int_{Q_T} |u(t,x)|^2\,dxdt<+\infty \}$

DEF2:$L^2(0, T; L^2(\Omega))=\{u: (0, T)\to L^2(\Omega), \mbox{measurable and} \int_{0}^T \|u(t)\|_{L^2(\Omega)}^2\,dt<+\infty \}$

We can easily identify the norms in the DEF1 and DEF2 by using Fubibi's theorem proved that $u: (0, T)\times \Omega \to \mathbb{R}$ is measurable. But if $u: (0, T)\times \Omega \to \mathbb{R}$ is not measurable, Fubini theorem can not be applied!

But my question focus on the measurability.

  1. If $u$ satisfies DEF1, hence $u$ is measurable on $(0, T)\times \Omega$ how can we prove that $u$ is a measurable on $(0, T)$ with values in $L^2(\Omega)$. Here $L^2(\Omega)$ is equipped the borel $\sigma$-algebra.

  2. If $u$ satisfies DEF2, hence $u$ is a measurable on $(0, T)$ with values in $L^2(\Omega)$. How can prove that $u$ is measurable on $(0, T)\times \Omega$.

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One of the problems here is that each element of your spaces is really an equivalence class of a.e. equal functions and not a genuine function. This is especially a problem in the case $u:(0,T)\to L^2$, because one has trouble selecting for each $t$ some representative function $x \mapsto u(t)(x)$.

One way to circumvent this is the following. Observe that the set $\Lambda$ of step functions that have "rectangles" as "steps" (i.e. sets of the form $A \times B$ with $A\subset (0,T)$ and $B\subset \Omega$ measurable) is dense in $L^2(\Omega_T)$.

Similarly, the set $\Theta$ of maps of the form

$$ t \mapsto \sum_{i=1}^n f_i \chi_{A_i} $$

with $f_i \in L^2(\Omega)$ and $A_i \subset (0,T)$ measurable is dense in $L^2((0,T);L^2(\Omega))$

Furthermore, the map

$$ \Gamma : \Theta \to L^2(Q_T), f \mapsto ((t,x)\mapsto f(t)(x)) $$

is well-defined and isometric (here, we use the special form of $\Theta$ and Fubini's theorem).

Hence, it extends to an isometric map $\tilde{\Gamma}: L^2((0,T),L^2(\Omega))\to L^2(Q_T)$. The isometry property implies that the image of $\tilde{\Gamma}$ is closed in $L^2(Q_T)$. But the image also contains the dense set $\Lambda$, so that $\tilde{\Gamma}$ is bijective (injective because of isometry).

It is now an easy exercise (using density) to verify that for each $f \in L^2((0,T);L^2(\Omega))$ there is a null-set $N_f \subset (0,T)$ such that $(\tilde{\Gamma}f)(t,\cdot)= f(t)$ holds for all $t \notin N_f$. Hence, $\tilde{\Gamma}$ yields the desired "measurable representative".

A similar property holds for the inverse map $\tilde{\Gamma}^{-1}$.

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