I'm trying to prove below result (that I'm not sure if it's true!). Could you have a check on my attempt?
Theorem: Let $E = F$ be non-empty sets. Let $|\cdot|, [\cdot]$ be norms on $E, F$ respectively such that they are equivalent. If $(E, |\cdot|)$ is reflexive, then so is $(F, [\cdot])$.
My attempt: Let $(E^{*}, |\cdot|_*), (F^{*}, [\cdot]_*)$ be the duals of $(E, |\cdot|)$ and $(F, [\cdot])$ respectively.
Equivalent norms induce the same topology and thus the same set of continuous linear functionals. Hence $E^{*} = F^{*}$. Notice that $|\cdot|, [\cdot]$ are equivalent if and only if $|\cdot|_*, [\cdot]_*$ are equivalent. This is due to $$ |x| = \max_{f \in E^*} \frac{f(x)}{|f|_*} \quad \text{and} \quad |f|_* = \sup_{x \in E} \frac{f(x)}{|x|}. $$
Let $(E^{**}, |\cdot|_{**}), (F^{**}, [\cdot]_{**})$ be the biduals of $(E, |\cdot|)$ and $(F, [\cdot])$ respectively. It follows from $|\cdot|_*, [\cdot]_*$ are equivalent that $|\cdot|_{**}, [\cdot]_{**}$ are equivalent. Again, we have $E^{**} = F^{**}$.
Let $J_E:E \to E^{**}$ and $J_F : F \to F^{**}$ be the canonical evaluation maps. We have $J_E$ is surjective. We have $$ \langle J_E x, f \rangle := f(x) \quad \text{and} \quad\langle J_F x, g \rangle := g(x) \quad \forall f \in E^*, \forall g \in F^*. $$
It follows that $J_E = J_F$ and thus $J_F$ is also surjective. This completes the proof.
