# Weak continuity of functions with values in a Banach space

I have problems understanding the proof of the following lemma:

Let $X$, $Y$ be Banach spaces, $X$ reflexive, and assume that $X$ is continuously, densely embeded into $Y$. Let $I \subset \mathbb{R}$ be open, bounded interval. Consider a function $\varphi \in L^{\infty}(I;X)$ such that $\varphi$ is weakly continuous from $\bar{I}$ to $Y$. $\space$ Then $\varphi$ is also weakly continuous from $\bar{I}$ to $X$.

In the proof, $I$ stands for not only the interval, but also for the mapping that represents the embedding from the assumptions of the lemma. Moreover, $I^{*}$ stands for the associated continuous, dense embedding of $Y^{*}$ to $X^{*}$.

I understand the proof until the point when they claim that $I \widetilde{\varphi}(t)=I\varphi(t)$ and $\widetilde{\varphi}(t)=\varphi(t)$. I don't see what it follows from.

I'll be thankful for any help.

I think the reason is because there is a typo. As you can notice, the special definition of $\tilde \varphi(t)$ is not really used (besides just being an integral), and the considered bounds are way too far off. When $\mu \in Y^*$, that limit is zero. My guess is the following, correct the definition by:

$$\left< \mathcal{J}(\tilde \varphi(t)), \mu \right>_{X^*} := \lim \inf_{h \rightarrow 0, t+h \in I} \frac{1}{h} \int_t^{t+h} \left< \mu, \varphi(s) \right>_X d\lambda_1(s)$$

In this case, when $\mu \in Y^*$, since $\varphi$ is weakly continuous in $Y$, $\left< \mu , I \varphi(s) \right>_Y$ is continuous. Therefore as a general property of continuous functions (I hope $d\lambda_1$ is the Lebesgue measure):

$$\lim \inf_{h \rightarrow 0, t+h \in I} \frac{1}{h} \int_t^{t+h} \left< \mu, I\varphi(s) \right>_Y d\lambda_1(s) = \left< \mu, I\varphi(t) \right>_Y$$

With this in mind, keep that $\mu \in Y^*$, we have that:

\begin{align} \left< \mu, I \tilde \varphi(t) \right>_Y = \left< I^* \mu, \tilde \varphi(t) \right>_{X} & = \lim \inf_{h \rightarrow 0, t+h \in I} \frac{1}{h} \int_t^{t+h} \left< I^* \mu, \varphi(s) \right>_X d\lambda_1(s) \\ & = \lim \inf_{h \rightarrow 0, t+h \in I} \frac{1}{h} \int_t^{t+h} \left< \mu, I\varphi(s) \right>_Y d\lambda_1(s) \\ & = \left< \mu, I\varphi(t) \right>_Y \end{align} Since this is true for all $\mu \in Y^*$, then $I \tilde \varphi(t) = I \varphi(t)$. In addition, since $$\left< I^* \mu ,\tilde \varphi(t) \right>_X = \left< I^* \mu, \varphi(t) \right>_X,$$ by density of $Y^* \subset X^*$, we conclude $\varphi(t) = \tilde \varphi(t)$.

I will not read the proof. Isn't it just the following: $\varphi$ has values in some bounded subset $B$ of $X$ which can be assumed to be absolutely convex and closed (= weakly closed). By reflexivity, $B$ is weakly compact and hence there is no strictly coarser Hausdorff topology on $B$. This implies that $\sigma(Y,Y^*)$ and $\sigma(X,X^*)$ coincide on $B$. As $\varphi$ is continuous w.r.t. the former topology it is continuous w.r.t. the latter.

• Thanks for your answer, it looks good. I just want to make sure that I understand why there is no strictly coarser Hausdorff topology on $B$: $B$ is weakly compact in $X$, hence it is also weakly compact in $Y$, but the weak mtopology on $B$ is Housdorff in both $X$ and $Y$, hence it is the same topology. Is it correct reasoning? – Jenda358 May 30 '18 at 15:29
• I think there's one problem: the above argument works only if $\varphi$ has values in $B$ at all times. Unfortunately, we know that only for almost all times. – Jenda358 May 30 '18 at 15:41
• You are right. That's a problem for the argument. – Jochen May 30 '18 at 15:59