Dear mathoverflowers, I have a question concerning the strong convergence in $L^p([0,T],X)$.

Let $X_1,X$ be two Banach spaces such that $X_1\subset X$ with compact embedding. Let $x_n(t)\in X_1$ be a bounded sequence in $X_1$ (this sequence converge strongly to $x(t)\in X$ for almost every $t\in [0,\infty)$). Moreover assume that $$ \|x_n(t)\|_X +\|x(t)\|_X \leq C $$ for almost every $t$.

Does this sequence converge strongly in $L^p([0,T],X)$? I think that, in fact, it does. I write my argument below. It seems quite easy, so I don't know if it is correct...

Let $T<\infty$ a fixed final time and define $\rho_n(t)=\|x_n(t)-x(t)\|^p_X$. Due to the compactness we have $\rho_n(t)\rightarrow 0$ for almost every $t$. We also have $\rho_n(t)\leq$C. I think that we can use now the dominated convergence theorem (taking $C$ as the dominating function and using the finiteness of the time interval) $$ \lim_n\int_0^T\rho_n(s)ds=0. $$

Therefore $$ \lim_n\|x_n-x\|_{L^p([0,T],X)}^p=\lim_n\int_0^T \|x_n(s)-x(s)\|^p_Xds=\lim_n\int_0^T\rho_n(s)ds=0, $$ and we obtain that $x_n$ converges strongly to $x$ in the Banach space $L^p([0,T],X)$.

Thank you in advance for your answer!!

  • $\begingroup$ Dear Rafa, it seems that some of your formulae is incomplete, something is missing.And, in particular, I miss your question... $\endgroup$ Commented May 18, 2013 at 12:33
  • $\begingroup$ Yes, you are right. I edited it properly. I don't know why the editor had problems with the sign '<'... $\endgroup$
    – guacho
    Commented May 18, 2013 at 14:03
  • $\begingroup$ Is the pointwise convergence $x_n(t)\rightarrow x(t)$ in $X$ a claim or an assumption? If it is a claim, it is false. $\endgroup$ Commented May 18, 2013 at 14:14
  • $\begingroup$ Why is $x_n(t)$ a measurable function of $t$? Is it an assumption? $\endgroup$
    – Hao Yin
    Commented May 18, 2013 at 14:17
  • $\begingroup$ I assume that $\|x_n(t)\|_{X_1}\leq M$ (M does not depend on $t$) for all positive times. Now fix $t$. Then, due to the boundedness, you can obtain $x_n(t)\rightarrow x(t)$ strongly in $X$. Of course, at this step, this convergence is pointwise a.e. $t$. Is it ok? $\endgroup$
    – guacho
    Commented May 18, 2013 at 14:38

1 Answer 1


Rafa: if I'm not mistaken your dominated convergence argument is wrong.

I agree that for a.e. fixed $t\in [0,T]$ you can extract a subsequence $n_k$ such that $x_{n_k}(t)\to x(t)$ strongly in $X$. The mistake you made is that the extraction procedure depends on the time $t$ you fix. Of course you can always use diagonal extraction to obtain $x_{n_k}(t)\to x(t)$ for a countable set of times $t$ (e.g. $t\in [0,T]\cap \mathbb{Q}$). But this is never going to be enough to obtain convergence $\rho_{n_k}(t)\to 0$ for a.e. $t\in[0,T]$, which is the hypothesis one needs to apply the Dominated convergence (in addition to suitable $L^p(0,T)$ bounds on $t\mapsto\rho_n(t)$, of course). In other words, the subsequence you extract must be chosen once and for all independently of the time $t$, which does not hold with only your hypotheses.

In a general way, you are missing compactness in time. Personally I like thinking of compactness in $L^p(0,T;X)$ as a different version of the Ascoli theorem: your compactness assumption $X_1\subset\subset X$ corresponds to the fact that the family $\{x_n(t)\}_n$ is pointwise relatively compact in $X$ for all time, but you are missing some kind of equicontinuity in time. A usual way to retrieve compactness in $L^{p}(0,T;X)$ is the Aubin-Lions lemma (see http://en.wikipedia.org/wiki/Aubin%E2%80%93Lions_lemma), which precisely takes care ot the time derivative and should be seen as some equicontinuity in time. Otherwise see also [Simon, " compact sets in the space $L^p(0,T;B)$", 1987]


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