My question concerns to the problem 6, chapter 7, of Evans PDE book (2nd edition).
In the book a hint is given but I couldn't get a solution from it. On the other hand, I got a solution without following the hint. So, I have two questions:
- How to solve the problem using the idea in the given hint?
- Is my solution (which doesn't follow the given hint) correct?
Problem: Suppose $H$ is a Hilbert space and $u_k\rightharpoonup u$ in $L^2(0,T;H)$. Assume further we have the uniform bounds $$\underset{0\leq t\leq T} {\operatorname{ess\;sup}}\|u_k(t)\|\leq C\qquad (k=1,2...) $$ for some constant $C$. Prove $$\underset{0\leq t\leq T} {\operatorname{ess\;sup}}\|u(t)\|\leq C$$
Given hint: We have $\int_a^b(v, u_k(t))\; dt \leq C\|v\||b - a|$ for $0 \leq a \leq b \leq T$ and $v\in H$.
My solution: I will use the identification $L^\infty(0,T;H)\cong [L^1(0,T;H')]'$.
The sequence $\{u_k\}$ is bounded in $L^\infty(0,T;H)$ because $$\|u_k\|_{L^\infty(0,T;H)}=\underset{0\leq t\leq T} {\operatorname{ess\;sup}}\|u_k(t)\|\leq C,\qquad \forall \ k\in\mathbb{N}.$$
It follows that there are a function $v\in L^\infty(0,T;H)$ and a subsequence of $\{u_k\}$ (which will not be relabeled) such that $$u_k \overset{*}\rightharpoonup v\quad\text{in}\quad L^\infty(0,T;H).$$ Therefore, $$\|v\|_{L^\infty(0,T;H)}\leq\liminf_{n\to\infty}\|u_k\|_{L^\infty(0,T;H)}\leq C.$$ But $$\left\{\begin{align} u_k\rightharpoonup u\quad&\text{in}\quad L^2(0,T;H)\\ u_k \overset{*}\rightharpoonup v\quad&\text{in}\quad L^\infty(0,T;H) \end{align}\right.$$ implies $u=v$ and thus $$\underset{0\leq t\leq T} {\operatorname{ess\;sup}}\|u(t)\|=\|u\|_{L^\infty(0,T;H)}=\|v\|_{L^\infty(0,T;H)}\leq C.$$
EDIT
Here are the details on the conclusion $u=v$. I'd like to know if it's ok.
We have $$\begin{align}u_k\overset{*}\rightharpoonup v\quad&\text{in}\quad L^\infty(0,T;H)\cong [L^1(0,T;H')]'\tag{1}\\ u_k\rightharpoonup u\quad&\text{in}\quad L^2(0,T;H)\cong [L^2(0,T;H')]'\tag{2} \end{align}$$
From $(2)$ we get $$u_k\overset{*}\rightharpoonup u\quad\text{in}\quad [L^2(0,T;H')]'\tag{3}$$ because $L^2(0,T;H')$ is reflexive. Precisely, $(1)$ and $(3)$ mean $$\left\{\begin{align} Tu_k\overset{*}\rightharpoonup Tv\quad&\text{in}\quad [L^1(0,T;H')]'\\ Fu_k\overset{*}\rightharpoonup Fu\quad&\text{in}\quad [L^2(0,T;H')]' \end{align}\right.$$ where $T$ is the isomorphism from $L^\infty(0,T;H)$ to $[L^1(0,T;H')]'$ and $F$ is the isomorphism from $L^2(0,T;H)$ to $[L^2(0,T;H')]'$. It follows that $$\begin{align} (Tu_k)(f)\to (Tv)(f)\quad&\text{in}\quad \mathbb{R},\qquad\forall\ f\in L^1(0,T;H')\tag{4}\\ (Fu_k)(h)\to (Fu)(h)\quad&\text{in}\quad \mathbb{R},\qquad\forall\ h\in L^2(0,T;H')\tag{5} \end{align}$$
From $(4)$ we get $$(Tu_k)(h)\to (Tv)(h)\quad\text{in}\quad \mathbb{R},\qquad\forall\ h\in L^2(0,T;H')\tag{6}$$
From the explicit form of the isomorphisms $T$ and $F$ we conclude that $$(Tu_k)(h)=(Fu_k)(h),\qquad\forall\ k\in\mathbb{N},\quad \forall\ h\in L^2(0,T;H')$$ So, $(5)$, $(6)$ and the uniqueness of the limit imply $$(Fu)(h)=(Tv)(h),\qquad\forall\ h\in L^2(0,T;H')\tag{7}$$ Again from the explicit form of the isomorphisms, we conclude that $$(Fv)(h)=(Tv)(h),\qquad\forall\ h\in L^2(0,T;H')\tag{8}$$ From $(7)$ and $(8)$ we get $$(Fu)(h)=(Fv)(h),\qquad\forall\ h\in L^2(0,T;H')$$ which implies $Fu=Fv$ which implies $u=v$.
