# Demonstrating that constructed solution to PDE satisfies initial condition

I'm having trouble filling in the details of a proof in Lions' book Quelques méthodes de résolution des problèmes aux limites non linéaires. He constructs approximate solutions to the wave equation on a bounded open set $\Omega\subset\mathbb{R}^n$ which satisfy the following conditions: \begin{align} u_n \rightarrow u \qquad&\text{weak-star in $L^\infty(0,T;H_0^1(\Omega)\cap L^p(\Omega))$} \\ u_n' \rightarrow u' \qquad&\text{weak-star in $L^\infty(0,T;L^2(\Omega))$} \\ \end{align} where $u_n(0)\rightarrow u_0$ strongly in $H_0^1(\Omega)\cap L^p(\Omega))$. (I have omitted some details to shorten the setup.)

I am having trouble with the claim on p. 14 that $u_n(0) \rightarrow u(0)$ weakly in $L^2(\Omega)$. Since, in particular, $u_n(0)\rightarrow u_0$ strongly in $L^2(\Omega)$, it follows that $u(0)=u_0$.

Attempt

First note that by interpolation, $u_n, u \in C([0,T];L^2(\Omega))$ for all $n\in\mathbb{N}$. Thus we interpret $u(0)$ as an $L^2$ limit, same for $u_n(0)$.

Let $\phi \in L^2(\Omega)$ be an arbitrary function and denote $$f_n(t) = \int_\Omega \phi (u_n-u) \, dx$$ and $\psi_m(t) = m*\chi_{[0,1/m]}(t)$. Then $$|f_n(0)| \leq \left|f_n(0) - \int_0^T \psi_m \, f_n(t) \, dt \right| + \left|\int_0^T \psi_m \, f_n(t) \, dt \right|.$$ For fixed $n$, the first term can be made small by the Lebesgue differentiation theorem (and noting the continuity of f). For fixed $m$, the second term can be made small by the weak convergence statement. But I couldn't get a diagonal sequence to conclude $f_n(0) \rightarrow 0$ as $n\rightarrow\infty$.

• Don't you have $u_n\to u$ in $L^\infty(0;T, L^2(\Omega))$ strongly? – daw Nov 25 '16 at 16:57
• By Aubin-Lions? Yes. – dls Nov 25 '16 at 18:06
• Then the second integral can be made small for large $n$ independent of $m$ – daw Nov 25 '16 at 19:22

By Aubin--Lions you have that $u_n\to u$ strongly in $L^\infty(0,T;L^2(\Omega))$, hence the second integral can be made arbitrarily small for large $n$ independent of $m$, as $f_n\to0$ in $L^\infty(0,T;L^2(\Omega))$.