Agreement of a pointwise, weak limit and a weak-$\ast$ limit of Bochner-integrable functions

Question

Let $T$ be a positive, real number and $\Omega \subset \mathbb{R}^d$ a bounded, connected, open set. For a given $p \in [2,\infty)$, I have a sequence of functions $(f_n)_{n \in \mathbb{N}} \subset L^\infty(0,T; L^p(\Omega))$, each of which is defined genuinely for every $t \in [0,T]$, not just a.e. $t$. Moreover, for each $n \in \mathbb{N}$ and $t \in [0,T]$, we have $f_n(t) \ge 0$ a.e. in $\Omega$ and $\int_\Omega f_n(t,x)\,dx = 1$, i.e., $(f_n(t))_{n \in \mathbb{N}}$ is a sequence of probability densities on $\Omega$. Finally, there exists some $M>0$, independent of $n \in \mathbb{N}$ and $t \in [0,T]$, such that $\| f_n(t) \|_{L^p(\Omega)} \le C$.

I can prove the following two statements:

1. By some information about where the $f_n$ come from, I can prove that there exists a function $g \colon [0,T] \to L^p(\Omega)$ that is weakly continuous in the sense that for every $t \in [0,T]$ and for every $\varphi \in L^{p'}(\Omega)$, where $p' := \frac{p}{p-1}$ is the Hölder conjugate of $p$, we have $\lim_{s \to t}\int_\Omega \big(g(t) - g(s)\big)\varphi \,dx = 0$. This function has the property that $g(t) \ge 0$ a.e. in $\Omega$ and $\int_\Omega g(t,x) \,dx = 1$. I can prove that, for every $t \in [0,T]$, we have $f_n(t) \rightharpoonup g(t)$ weakly in $L^p(\Omega)$.
2. From the uniform bound $\sup_{n \in \mathbb{N}} \sup_{t \in [0,T]} \| f_n(t) \|_{L^p(\Omega)} \le M$, there exists $h \in L^\infty(0,T; L^p(\Omega))$ such that $f_n \rightharpoonup h$ weakly-$\ast$ in $L^\infty(0,T; L^p(\Omega))$.

My question: is it true that for a.e. $t \in [0,T]$ we have $g(t) = h(t)$?

I'm not sure how to begin proving or disproving this. Can someone please provide some guidance? Thank you!

Answer 1

Take $\phi \in C_c^\infty(0,T)$ and $v\in L^q(\Omega) = L^p(\Omega)^*$. Then $\phi v \in L^1(0,T;L^{p'}(\Omega))$, so that $$ \int_0^T \int_\Omega (f_n-h) v \phi dx\ \ dt \to 0. $$ For fixed $t$, the function $t\mapsto \int_\Omega f_n(t) v(x) \phi(t)\ dx$ converges pointwise to $ \int_\Omega g(t) v(x) \phi(t)\ dx$. In addition, we have the pointwise bound $$ \left| \int_\Omega f_n(t) v(x) \phi(t)\ dx\right| \le \|f_n\|_{L^\infty(0,T;L^p(\Omega))} \|v\|_{L^1(0,T;L^{p'}(\Omega))} \|\phi\|_{L^\infty}, $$ which is integrable. Using dominated convergence theorem, we can pass to the limit $$ \int_0^T \int_\Omega f_n(t) v(x) \phi(t)\ dx \ dt \to \int_0^T \int_\Omega g(t) v(x) \phi(t)\ dx \ dt . $$ Since $\phi$ is arbitrary it follows $\int_\Omega (g(t)-h(t))v(x) \ dx=0$ for almost all $t$. Since $v$ is arbitrary, it follows $g=h$ a.e.