Question about Lebesgue Bochner spaces

Question

Let $T>0$ and $\Omega\subset\mathbb{R}^N$ be a bounded domain. Also $p\in (1,\infty)$ is any number.

I know that $u\in L^{p}((0,T);L^p(\Omega))$ and $\nabla u\in L^{p}((0,T);L^p(\Omega))^N$. How can I show that $u\in L^{p}((0,T);W^{1,p}(\Omega))$?

It is easy to show that $\displaystyle\int_{0}^T \Vert u(t,\cdot)\Vert_{W^{1,p}(\Omega)}^p\ dt<\infty$. I'm having problems in showing that $u:(0,T)\to W^{1,p}(\Omega)$ is Bochner measurable.

So we need to find a sequence $u_n:(0,T)\to W^{1,p}(\Omega)$ of simple functions such that $\Vert u_n(t)-u(t)\Vert_{W^{1,p}(\Omega)}\longrightarrow 0$ for a.a. $t\in (0,T)$.

How can we do that?

Simple functions that are obtained as sum of characteristic functions of cartesian products $A\times B$ with $A\subset (0,T)$ and $B\subset\Omega$ are dense in $L^{p}((0,T)\times\Omega)$ (see here Simple functions on a product measure space)...I am stucked here.

Answer 1

By Pettis' theorem, $u \colon (0,T) \to W^{1,p}(\Omega)$ is strongly [Bochner] measurable if and only if it is weakly measurable due to separability of $W^{1,p}(\Omega)$.

It is thus sufficient to show that $t \mapsto \langle \Phi, u(t)\rangle$ is measurable $(0,T) \to \mathbb R$ for any $\Phi \in W^{1,p}(\Omega)^*$, the dual space of $W^{1,p}(\Omega)$. Then again, any such $\Phi \in W^{1,p}(\Omega)$ can be represented by a pair $f \in L^q(\Omega)$ and $F \in L^q(\Omega)^n$, with $\frac1p + \frac1q = 1$, via $$\langle \Phi, u(t)\rangle = \int_\Omega f\cdot u(t) + \int_\Omega F \cdot \nabla u(t).$$ This is a standard result in Sobolev spaces, see for example Proposition 8.14 in Brezis.

In the present case, from the assumptions that $u \colon (0,T) \to L^p(\Omega)$ and $\nabla u \colon (0,T) \to L^p(\Omega)$ are measurable, both integrals in the foregoing expression for $\Phi$ are measurable $(0,T) \to \mathbb R$, due to $L^q(\Omega) = L^p(\Omega)^*$, and thus so is $t \mapsto \langle \Phi, u(t)\rangle$.

It follows that $u \colon (0,T) \to W^{1,p}(\Omega)$ is weakly and thus, equivalently, strongly measurable. (I concede that this does not answer the precise question asked in OP about approximation by simple functions, but solves the original problem.) Note also:

• The whole argument also works for $p=1$ as far as I see.
• The representation of $\Phi$ by $f$ and $F$ is non-unique.
• The argument one uses to derive the representation for functionals in $W^{1,p}(\Omega)^*$ goes by considering $W^{1,p}(\Omega)$ (isometrically) as a closed subspace of $L^p(\Omega)^{n+1}$ via $u \mapsto (u,\nabla u)$. Probably one could also derive measurability of $u \colon (0,T) \to W^{1,p}(\Omega)$ from that.