How do you prove the following?
If $\mathcal{X}$ is reflexive and $M \leq \mathcal{X} \rightarrow \mathcal{X}/M$ is reflexive
There is no assumption that $\mathcal{X}$ is a Banach space.
How do you prove the following?
If $\mathcal{X}$ is reflexive and $M \leq \mathcal{X} \rightarrow \mathcal{X}/M$ is reflexive
There is no assumption that $\mathcal{X}$ is a Banach space.
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$\mathcal X$ reflexive, so every functional on $\mathcal X^*$ is an evaluation map. Now consider a functional $g$ on $(\mathcal X/M)^*$. One identifies $(\mathcal X/M)^*$ with the functionals on $\mathcal X$ that vanish on $M$. Now, extend $g$ to all of $\mathcal X^*$ by declaring it $0$ outside of $(\mathcal X/M)^*$. Now we get $x\in\mathcal X$ such that $g$ is the map $f\mapsto f(x)$
Then, restrict $g$ to just the functionals that vanish on $M$.
This gives rise to the realisation of $g$ as an evaluation map.
Since $g$ was arbitrary. We're done $\square$