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I would like to know if this statement is true. If so would you like to know how to prove it or some counter-example in the negative case.

Let $I$ be a interval in $mathbb{R}$ (not necessarily limited), $H'$ be the dual of a Hilbert space H and consider $(u_n) subset L^{infty}(I;H')$ such that $u_n overset{ast}{rightharpoonup} u $ in $L^{infty}(I;H')$, then $u_n rightharpoonup u$ in $L^2_{operatorname{loc}}(I;H').$

In the case that I am considering $H=L^2(Omega)$, which is being used as pivot space in the chain $H_{0}^{1}(Omega) hookrightarrow H=H' hookrightarrow H^{-1}(Omega)$, where $Omega$ is a bounded domain of $mathbb{R}^{d}$ with regular boundary.

What I thought was based on the following questions:
The dual space of locally integrable function space, Dual space of Bochner space and
Compute the dual of the $L^1$ space of $L^1$-valued functions (Lebesgue-Bochner space).

Affirmations:

i) $(L_{operatorname{loc}}^{2}(I;H')'=L^2_{operatorname{comp}}(I;H)$.

ii) The application $overline{v} mapsto v$ from $L^1(I;H)'$ into $L^infty(I;H')$ where $langle overline{v}, u rangle = int_{I} langle v(t),u(t) rangle dt$ for all $u in L^1(I,H)$ is bijective and $|overline{v}|=|v|_{infty}$.

iii)The convergence $u_n overset{ast}{rightharpoonup} u $ in $L^{infty}(I;H')$ is characterized by $int_{I} langle u_{m}(t), v(t) rangle_{H',H} dt rightarrow int_{I} langle u(t), v(t) rangle_{H',H} dt$.

Note that the last two conditions are fulfilled when I is limited. (Zeidler, E. Nonlinear Functional Analysis and Its Applications II-A., Exercise 23.12)

Let $v in (L^2_{operatorname{loc}}(I;H')'= L^2_{operatorname{comp}}(I;H) hookrightarrow L^1(I; H)$, since $v$ is zero outside of a compact subset of $I$. Then,

$$langle v,u_n rangleequiv int_{I} langle u_{m}(t), v(t) rangle_{H',H} dt rightarrow int_{I} langle u(t), v(t) rangle_{H',H} dt= langle v,urangle.$$

By the above URL, it seems that dual space of $L^1(I;H)$ is not $L^infty(I;H')$, when $I$ is not bounded.

Is there another way to prove it?