$L^{2}(0,T; L^{2}(\Omega))=L^{2}([0,T]\times\Omega)$? Suppose I have a function $u\in L^{2}([0,T]\times\Omega)$ for some bounded domain $\Omega$ in $\mathbb{R}^{2}$ or $\mathbb{R}^{3}$. I managed to prove that this implies $u\in L^{2}(0,T; L^{2}(\Omega))$. Now I want to show that opposite is not true. This is where I struggle.
According to Fubini's theorem, it is sufficient to give example of a function $u\in L^{2}(0,T; L^{2}(\Omega))$ which violates condition $u\in L^{2}(\Omega; L^{2}(0,T))$.
So could you please provide such function?
Thanks to all!
Let me show you how I see space $L^{2}(0,T; L^{2}(\Omega))$. The problem with it is that I couldn't find a single definition in any book whose rigorousness satisfied me.
So, as far as I understand, $L^{2}(0,T; L^{2}(\Omega))$ is a set of functions $\{f : [0,T]\times\Omega\to\overline{\mathbb{R}}|\text{for almost all } t\in[0,T] f(x,t)\in L^{2}(\Omega)|\int_{0}^{T}\int_{\Omega}f^{2}dxdt < +\infty\}$ and such that for any pair of functions $f, g$ from this set the function $\int_{\Omega}fgdx$ is measurable in time. The last condition is very crucial and is omitted everywhere like it's obvious or something. Note that condition that the integral is finite is well-defined, since $f^{2}$ is non-negative and to define Lebesgue integral for it you do not even have to make $f^{2}$ measurable. Also, integral $\int_{\Omega}fgdx$ is also well-defined for alsmot all $t$, since the integral of the absolute value is bounded via Holder's inequality and $f, g$ are assumed measurable since they are in $L^{2}$, so $\int_{\Omega}fgdx$ exists and is finite.
It troubles me that they keep using both $L^{2}([0,T]\times\Omega)$ and $L^{2}(0,T; L^{2}(\Omega))$ in literature. If these spaces are indentical, why different notations? So I assumed that $L^{2}(0,T; L^{2}(\Omega))$ is wider than $L^{2}([0,T]\times\Omega)$, which is why I ended up here at stackexchange.
My main concern is with PDEs. So, suppose I have some time-dependent second-order PDE in a form $Lu = f$. To make it simple, assume $u = 0$ on the boundary of $\Omega$ and $u(0) = 0$, $u_{t}(0) = 0$. You can think of $L$ as a wave operator. Now, if we multiply the equation by test function and integrate over entire $[0,T]\times\Omega$, we obtain some weak formulation. But we can instead integrate over $\Omega$ only and assume the integral relation holds for almost all $t$. Naturally, the test functions in both cases will be from different spaces, so are solutions we are looking for. One is, say, $L^{2}([0,T]\times\Omega)$, the other is $L^{2}(0,T; L^{2}(\Omega))$. So, the question is: if in both formulations we avoid all sorts of integrations by parts ( or do them similarly ), will we obtain the same existence-uniqueness results for $u$, as well as the same regularity results? I searched tons of literature, and looks like the only formulation people are interested in is the one with entire integration over $[0,T]\times\Omega$. But they keep using the second one when obtaining some a priori estimates in the papers devoted to Finite Element approximations. Which really pisses me off, since they don't bother proving existence and uniqueness for the second formulation.
 A: Without any literature I am not sure that everything I write down, is easily to understand. However this can bring some light into your thoughts...:-)
I directly follow the proof of Etienne Emmrich in his book 
"Gewoehnliche und Operator-Differentialgleichungen".
Theorem: Let $1\leq p<\infty$ be given and $a,b\in\mathbb{R}$. Then we have
$$ L^{p}(0,T;L^{p}(a,b))\cong L^{p}((a,b)\times (0,T)).$$
For a proof of this result we need the following Lemma:
Lemma: Let $\tilde{u}=\tilde{u}(t):[0,T]\rightarrow L^{p}(a,b)$ be Bochner measurable $(1\leq p\leq\infty)$. Then the real valued function $u=u(x,t):=[\tilde{u}(t)](x)$ is Lebesgue measurable over $(a,b)\times (0,T)$.
Proof: Since $\tilde{u}$ is Bochner measurable there exists a sequence of simple functions $\{\tilde{u}_{n}\}$ where we have
$$\int_{a}^{b}|u_{n}(x,t)-u(x,t)|^{p}\rightarrow 0\ \hbox{ for } n\rightarrow\infty$$
and for the case $p=\infty$ the analogous result with the $L^{\infty}$-norm
for $t\in[0,T]$ a.e. Thereby we used $u_{n}(x,t):=[\tilde{u}_{n}(t)](x)$. Then there exists a subsequence, such that for $x\in[a,b]$ a.e. and $t\in[0,T]$ a.e. 
we otain
$$|u(x,t)-u_{n'}(x,t)|\rightarrow 0\ \hbox{ for } n'\rightarrow\infty.$$
The functions $u_{n'}$ are measurable over $(a,b)\times(0,T)$ by construction, since the have only finitely many values with respect to $t$ and for every $t$ they are Lebesgue measurable with respect to $x$.
So we constructed a sequence of Lebesgue measurable functions converging a.e. to the function $u$. hence $u$ is Lebesgue measurable. $\square$
Let us come back to our "main" Theorem. We give the following proof:
Firstly we show $$L^{p}(0,T;L^{p}(a,b))\supseteq L^{p}((a,b)\times(0,T)).$$
For that let $u\in L^{p}((a,b)\times(0,T))$ be given. We define the "abstract" functio  $\tilde{u}$ by $[\tilde{u}(t)](x):=u(x,t)$. Since $u(\cdot,t)\in L^{p}(a,b)$ we also get $\tilde{u}$ maps into $L^{p}(a,b)$ as well. 
Now pick $f\in (L^{p}(a,b))^{*}\cong L^{q}(a,b)$ ($\tfrac{1}{p}+\tfrac{1}{q}=1 \hbox{ and } q=\infty\hbox{ for } p=1$). Then 
$$(x,t)\mapsto f(x)u(x,t)\in L^{1}((a,b)\times (0,T))$$ and by Fubini's Theorem the map
$$t\mapsto \int_{a}^{b}f(x)u(x,t)\ dx =\langle f,\tilde{u}(t)\rangle$$
on $[0,T]$ Lebesgue integrable such that also $t\rightarrow \langle f,\tilde{u}(t)\rangle$ Lebesgue measurable.
This is the weak measurability of $\tilde{u}$. Since $L^{p}$ is seperable for $1\leq p<\infty$ we have the Bochner measurability of $\tilde{u}$. $\tilde{u}$ is also Bochner integrable since by Fuibini's theorem also the map
$$t\rightarrow \int_{a}^{b}|u(x,t)|^{p}\ dx=\|\tilde{u}(t)\|^{p}_{L^{p}}$$
is Lebesgue integrable on $[0,T]$.
It remains to show $$L^{p}(0,T;L^{p}(a,b))\subseteq L^{p}((a,b)\times(0,T)).$$
Let $\tilde{u}\in L^{p}(0,T;L^{p}(a,b))$. We again define the real valued function $u(x,t):=[\tilde{u}(t)](x)$. The iterated integral
$$
\int_{0}^{T}\left(\int_{a}^{b}|u(x,t)|^{p}\ d x\right) \ dt<\infty
$$
exists. Since the function to integrate over is non negative, we can conclude that also
$$\int_{(a,b)\times(0,T)}|u(x,t)|^{p}\ d (x,t)$$ exists and is finite, if and only if $u$ is Lebesgue measurable on $(a,b)\times(0,T)$. This is the result of the aforementioned lemma. Finally 
$$
\int_{0}^{T}\left(\int_{a}^{b}|u(x,t)|^{p}\ d x\right) \ dt=\int_{(a,b)\times(0,T)}|u(x,t)|^{p}\ d (x,t)
$$ and thus the norms are identical. $\square$
