Suppose $H$ is Hilbert space and $u_k$ converges weakly to $u$ in $L^2(0,T;H)$.

Suppose further we have the uniform bounds

$\mathrm{esssup}_{0≤t≤T} ||u_k(t)||≤C$.

Then $\mathrm{esssup}_{0≤t≤T} ||u(t)||≤C$.

I cannot prove this question. I think that $u_k(t)$ converges weakly to $u(t)$ for every $t$, but I cannot. Please tell me this question.

  • $\begingroup$ Can you prove this follow the hint in Evans'book now? $\endgroup$ – Kira Yamato Jan 16 '15 at 6:13
  • $\begingroup$ @Ylath Take the limit $k\to\infty$ in the hint and use $u=v$ as test function. Now use $||u||=\sup_{||v||\leq 1}|(u,v)|$ and conclude. $\endgroup$ – alemou Jan 18 '15 at 11:23
  • $\begingroup$ @alemou I still don't know how to use that hint, can you give an answer below? Thank you very much, I think so many people want to know. $\endgroup$ – Kira Yamato Jan 18 '15 at 11:34

Hint: by Mazur's lemma, we can pick a convex combination $v_n$ of $u_k$ such that $v_n$ converges strongly to $u$. Then there is a subsequence of $v_n$ which converges to $u$ almost everywhere and it's easy to see $v_n$ satisfies the uniform bounds, so is its almost everywhere limit


Evan's hint:

For all $v\in H$ $$ \int_0^T (v,u_k)\leq C||v|| T. $$

Since $L^2(0,T;H)$ is Hilbert, the assumption $u_k \rightarrow u$ weakly in $L^2(0,T;H)$ reads \begin{equation} \int_0^T (v,u(t)) dt=\lim_{k\to\infty}\int_0^T(v,u_k(t)), \,\,\forall v\in L^2(0,T;H) \end{equation}

Consider the particular case, $v\in L^2(0,T;H)$ as $v= w$ with $w\in H$ non-depending on time. Using the hint $$ \int_0^T( v,u_k(t))dt\leq ||v|| C T $$ Take the limit for $k\to\infty $ in the l.h.s. $$ \int_0^T(v,u(t))dt\leq||v||C T, \,\,\forall v\in H $$ Take now on both members of the last inequality the supremum on $||v||\leq 1$, we get $$ \int_0^T ||u(t)||dt=\int_0^T\sup_{||v||\leq 1}(v,u(t))\leq CT $$ From which $||u(t)||\leq C$ for all $t\in [0,T]$, thus $\text{ess}\sup_{[0,T]}||u||\leq C.$

  • $\begingroup$ Can you explain more about the final step? why is the integration <CT deduce norm <C $\endgroup$ – Kira Yamato Jan 18 '15 at 11:55
  • $\begingroup$ I have edited the answer, it's still the same argument. you have $\int_0^T||u||\leq CT$, then $\int_0^T(||u||-C)dt\leq 0$, and thus $||u||\leq C$ for all t $\endgroup$ – alemou Jan 18 '15 at 12:18

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