How does the space $L^2(\Omega_T)$ compares to $L^2(0,T; L^2(\Omega))$?

Question

Let $\Omega \subset \mathbb{R}^n$ and $\Omega_T = \Omega \times (0,T)$.

I am confused with regards to the difference between the space $L^2(\Omega_T)$ and $L^2(0,T; L^2(\Omega))$. Are they equivalent? The norm for the latter is

$$\int_0^T \|u\|_2^2 = \int_0^T \int_{\Omega} |u|^2$$

I am not sure what will be the norm for the former (Although the domain will be $\Omega_T$)

Answer 1

Edit. I corrected some errors that were misleading.

The norm of the first space, $L^2(\Omega_T)$, is $$ \|f\|_{L^2(\Omega_T)}:=\left(\int_{\Omega_T} |f(x,t)|^2 dx\,dt\right)^{1/2}. $$ It is simply the space of $L^2$ functions on the measure space $\Omega_T$ seen as a subset of $\mathbb R^{n+1}$. Basically the usual Lebesgue space.

The answer to your question depends a little (although not much) on how the second space $L^2(0,T;L^2(\Omega))$ is defined:

1. If it is defined as the space of measurable functions on $\mathbb \Omega_T$ such that the norm on the right hand side of your formula is finite, then it coincides trivially with $L^2(\Omega_T)$ thanks to Tonelli's theorem.

2. There is another canonical way of defining $L^2(0,T;L^2(\Omega))$, and that is in the setting of Bochner spaces: basically, you define first what is a measurable function from $[0,T]$ to any Banach space $X$, and then you consider the space $L^p(0,T;X)$ of all the measurable functions from $[0,T]$ to $X$ such that the norm on the right hand side of your equation is finite. In that case, the two spaces are formally two different kinds of objects, but they are still morally the same thing, and in fact one can prove that they are canonically equivalent, i.e., isometrically isomorphic (the proof of course needs a good understanding of what is a Bochner space). It goes without saying: any function from $[0,T]$ to a space of functions defined on $\Omega$ can be formally seen as a function of $n+1$ variables, this is the identification between the two spaces.

So in both cases, they pretty much coincide.

My answer is: if you only care about what the second space is in practice (like for PDE applications, etc.), it is basically just $L^2(\Omega_T)$. If your problem is trying to understand more rigorously what that space is and how it is defined in a specific paper and you need to dig a little deeper into the details of real analysis, then you should learn a little bit about Bochner spaces. Most of PDE papers and books do not even care about defining these spaces (although generally, when people write $L^2(a,b;X)$, they likely mean the Bochner space) because all the definitions you can imagine are essentially equivalent and the subtleties involved are generally of no particular interest for applications in PDEs, so you can often safely ignore all of this and know that it's simply $L^2(\Omega_T)$ written in a different form. But for research purposes, and in general if you care about understanding the details, it is definitely worth to know the basics of the theory of Bochner spaces, because for some rigorous proofs you might need to prove something using that formalism.

Answer 2

$L^2(\Omega_T) = L^2(\Omega_T ; \mathbb{C})$ is the space of square integrable functions from $\Omega_T$ to $\mathbb{C}$, with equality defined by a.e. equivalence. $L^2(0, T ; L^2(\Omega)) = L^2([0, T] ; L^2(\Omega ; \mathbb{C}))$ is the space of square integrable functions from $[0, T]$ to $L^2(\Omega)$ with equality defined by a.e. equivalence. For the definition of the integral of $f \in L^2(0, T ; L^2(\Omega))$, known as the Bochner integral, see https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/vecint.pdf

Formally, ny Tonelli's theorem, the norms on both spaces are the same, but I'm not sure whether the spaces are the same.