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I would like to know the proof of the following theorem:

Consider two Banach spaces $X\hookrightarrow Y$ and $1<p,q\leq\infty$. Let $(f_n)_{n\geq 0}$ be a bounded sequence in $L^q(I,Y)$ and let $f:I\mapsto Y$ be such that $f_{n}(t)\rightharpoonup f(t)$ in $Y$ for a.a. $t\in I$. If $f_n$ is bounded in $L^p(I,X)$ and if $X$ is reflexive, then $f\in L^p(I,X)$ and $\|f\|_{L^p(I,X)}\leq\liminf_{n\rightarrow}\|f_n\|_{L^p(I,X)}$.

It is stated as Theorem 1.2.5 in Cazenave's Semilinear Schroedinger Equations but is given without proof.

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  • $\begingroup$ I guess that $p<\infty$, right? $\endgroup$ Mar 17, 2014 at 17:22
  • $\begingroup$ Maybe you're right. I've just copied the theorem from the book of Cazenave:semilinear Schrödinger equation. $\endgroup$ Mar 17, 2014 at 17:26
  • $\begingroup$ The keyword you are looking for is '(the) Direct Method in(of) the Calculus of Variations' I think. $\endgroup$
    – username
    Mar 18, 2014 at 9:31

2 Answers 2

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Here is an answer when $p<\infty$ and $I$ bounded. The argument might be adapted to $p=\infty$ by replacing some weak $L^p$ convergences below by the weak-* $L^{\infty}$ one, and I really believe $I$ bounded is not an issue (otherwise argue locally on any $J\subset I$)

Just a few obvious remarks first: if $X$ is reflexive and $f_n$ bounded in $L^p(I,X)$ with $p<\infty$ (resp. $p=\infty$) then by the Banach-Alaoglu theorem we can extract a subsequence $f_{n_k}\rightharpoonup \tilde{f}$ in this topology (resp. $f_{n_k}\overset{*}{\rightharpoonup} \tilde{f}$). Also, $X\hookrightarrow Y$ implies $ Y'\hookrightarrow X'$.

  1. Claim: $f_n\rightharpoonup f$ in $L^p(I,Y)$. To see this, fix a test function $\varphi\in L^{p'}(I,Y')$ and let $$ h_n(t):=\left<f_n(t),\varphi(t)\right>_{Y,Y'}. $$ By the OP's assumptions we have $$ h_n(t)\to h(t)=<f(t),\varphi(t)>_{Y,Y'}\quad \text{a.e. }t\in I, $$ but also since $f_n\in L^{p}(I,X)\Rightarrow f_n(t)\in X$ a.e. and $Y'\hookrightarrow X'$ $$ h_n(t)=\left<f_n(t),\varphi(t)\right>_{X,X'}\in L^1(I). $$ My claim would immediately follow from $h_n\to h$ in $L^1(I)$, which we can get by Vitali's convergence theorem. In order to apply the latter we only need to check: (i) that $|I|<\infty$ (OK since I'm considering bounded $I$), (ii) $h_n(t)\to h(t)$ for a.e. $t\in I$ (OK by the OP's assumption), (iii) that $|h(t)|<\infty$ for a.e. $t$ (still OK because $f_n(t)\rightharpoonup f(t)$ in $Y$ and the test function is fixed so $\varphi(t)\in Y'$ for a.e. $t$), and finally (iv) that $\{h_n\}_{n}$ is uniformly integrable. The only delicate point is to check (iv), which by the Dunford-Pettis is equivalent to showing that $\{f_n\}_{n}$ is relatively compact for the weak $\sigma(L^1,L^{\infty})$ topology. In order to check this relative compactness, recall from my preliminary remark that $f_{n_k}\rightharpoonup \tilde{f}$ in $L^{p}(I,X)$. For any fixed $\lambda\in L^{\infty}(I)$ we have $\lambda\varphi\in L^{p'}(I,Y')\hookrightarrow L^{p'}(I,X')$ so along this subsequence \begin{align*} \int_I h_{n_k}(t)\lambda(t) & =\int_I\left<f_{n_k}(t),\varphi(t)\right>_{X,X'}\lambda(t)\\ & =\int_I\left<f_{n_k}(t),\lambda\varphi(t)\right>_{X,X'}\\ &\underset{k\to\infty}{\to} \int_I\left<\tilde{f}(t),\lambda\varphi(t)\right>_{X,X'}\\ & =\int_I \tilde{h}(t)\lambda(t) \end{align*} with $\tilde{h}(t)=\left<\tilde{f}(t),\varphi(t)\right>_{X,X'}$. This means that $\tilde{h}\in L^{1}(I)$ is a cluster point for the $\sigma(L^1,L^{\infty})$ topology. Thus $\{h_n\}_{n}$ is weakly $L^1$ relatively compact hence also uniformly integrable, and my claim follows.

  2. Improvement to $L^p(I,X)$. From step 1 and my preliminary remark we have now $$ f_n\rightharpoonup f\text{ in }L^{p}(I,Y),\qquad f_{n_k}\rightharpoonup \tilde{f}\text{ in }L^{p}(I,X). $$ Note that the whole sequence converges for the weak $Y$ norm, but only up to a subsequence for the stronger $X$ norm. Restricting to test-functions $\varphi\in L^{p'}(I,Y')\subset L^{p'}(I,X')$ and by uniqueness of the limit it follows that $f=\tilde{f}$ in $L^{p}(I,Y)$. In particular $f\in L^{p}(I,X)$, and because $\tilde{f}$ was obtained extracting a weakly $L^p(I,X)$ converging subsequence we get $$ ||f||_{L^p(I,X)}=||\tilde{f}||_{L^p(I,X)}\leq \liminf\limits_{n\to\infty} ||f_n||_{L^p(I,X)} $$ as desired and the proof is complete.

Note that I never used the $L^q(I,Y)$ bound, so this assumption can probably be removed. The pointwise convergence in $Y$-weak a.e. $t$ is a quite strong assumption to start with, so this is not surprising.

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I also had troubles with this theorem when I read the Cazenave's book. Recently I found this answer for proving the theorem. However, I think there is still a little flaw in the proof, namely in step 1, we can not directly use that $L^p(I,Y)$ has dual $L^{p'}(I,Y')$, since this is valid for instance under the condition that $Y$ has Radon-Nikodym-property. So I would like to write up my proof and I hope that everything is ok. The basic idea is essentially the same as the one given in the originial proof, but with a minor modification.

WLOG we can assume that $|I|<\infty$, for general case we can show the claim by using monotone convergence theorem. We also restrict ourselves to the case $1<p<\infty$, for $p=\infty$ we can just replace the weak topology arguments by the corresponding weak-$*$-arguments. Since $X$ is reflexive, we know that $X$ has RNP, thus together with the boundedness of $(f_n)_n$ in $L^p(I,X)$ we know that $f_n$ converges weakly to some $h$ in $L^p(I,X)$. Now let $m$ be a number in the interval $(1,\min(p,q))$. From the assumption that $f_n$ is bdd in $L^q(I,Y)$ and the fact that $|I|<\infty$ we know that $f_n$ is bdd in $L^m(I,Y)$, thus using Fatou's lemma one deduces that $f$ is also in $L^m(I,Y)$. Now let $\phi\in L^{m'}(I,Y')$. we want to show that $$\int_{I}\left<f_n(t)-f(t),\phi(t)\right>_{Y,Y'}dt\to 0 $$ as $n\to\infty$ (Note I am not showing duality! Since $Y$ does not have RNP). To see this, since $|I|<\infty$ and the integrand in the above integral converges to zero a.e. (which is shown in the original answer), we can use Egoroff's theorem to decompose $I$ into $K$ and $K^c$ such that the integrand converges to zero uniformly on $K$, while the measure of $K^c$ is small. Using absolute continuity of the integral of an integrable function we can make the integral involving $K^c$ arbitrary small, as long as the measure of $K^c$ is small, which is possible. On the other hand, due to embedding we know that $f_n$ converges to $h$ in $L^m(I,Y)$ weakly, thus we have (notice that $L^{m'}(I,Y')$ is a subspace of $(L^m(I,Y))'$, as long as $Y$ is a Banach space and $m\in[1,\infty)$) $$\int_{I}\left<f(t)-h(t),\phi(t)\right>_{Y,Y'}dt=0 $$ for all $\phi\in L^{m'}(I,Y')$. Using fundamental lemma of calculus of variations we conclude that $f=h$ a.e., and hence $f$ is in $L^{p}(I,X)$, and $f_n$ converges weakly to $f$ in $L^p(I,X)$. The last statement can now be deduced by the lower semicontinuity of the norm function.

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