Suppose $U$ is a Banach space and let $V\subseteq U$ be bounded, convex and (norm-)closed. Consider the Bochner-Lebesgue space $L^r(0,T;U)$ with $T>0$ and $r\in[1,\infty]$ consisting of strongly measurable functions $u\colon[0,T]\to U$ with $\int_0^T\Vert u(s)\Vert^r_U\,ds<\infty$. For such functions $u$, we define the integral average $w_\varepsilon(s)=(T_\varepsilon u)(s)=\frac1\varepsilon\int_s^{s+\varepsilon}u(t)\,dt$, whenever $s+\varepsilon<T$.
Now, given a net $(u_\iota)_\iota$, weakly$^\ast$ converging in $L^r(0,T;U)$ to some $u$ and being a.e. $V$-valued for any $\iota$, what can we say about convergence of $w_{\varepsilon,\iota}=T_\varepsilon u_\iota$, hopefully against $w_{\varepsilon}=T_\varepsilon u$?
My attempts so far:
Naively inserting and calculating: For any $\varphi\in L^{r/(r-1)}(0,T;U_\#)$ (we view $L^r(0,T;U)$ as a dual space), we have $$ \langle u_\iota-u,\varphi\rangle_{L^r,L^{r/(r-1)}}=\int_0^T\langle (u_\iota-u)(s),\varphi(s)\rangle_{U,U_\#}\,ds\xrightarrow[]{\iota}0 $$ by assumption. Assuming, that $w_{\varepsilon,\iota},w_\varepsilon\in L^r(0,T;U)$ as well, $$ \langle w_{\varepsilon,\iota}-w_\varepsilon,\varphi\rangle_{L^r,L^{r/(r-1)}}=\int_0^T\langle (w_{\varepsilon,\iota}-w_{\varepsilon})(s),\varphi(s)\rangle_{U,U_\#}\,ds=\int_0^T\frac1\varepsilon\int_s^{s+\varepsilon}\langle(u_\iota-u)(t),\varphi(s)\rangle_{U,U_\#}\,dt\,ds, $$ which I so far didn't seem to get working.
Assuming a stronger type of convergence: If $\Vert u_\iota-u\Vert_{L^\infty(0,T;U)}\to0$, then $w_{\varepsilon,\iota}-w_{\varepsilon}\to0$ strongly and weakly$^\ast$. This only covers the case $r=\infty$ and assumes strong convergence, which I don't think I get from my context.
We can adjust $r$, if needed, and perhaps polish my definition of averages a bit. I saw other posts (e.g. Average on the whole space) that relate to my question. Any help is appreciated. Thanks, Joe
