= Solution
If $X$ is a <reflexive Banach space>, its closed unit ball is weakly compact. A bounded linear map is weak-to-weak continuous, so $T(B_X)$ is weakly compact in $Y$. Compact subsets of a Hausdorff space are closed, hence $T(B_X)$ is weakly closed and therefore norm closed.
Now suppose the two norms on $V$ have the same continuous dual as a set. Each $X_j^*$ is a <Banach space> in its dual norm. The identity
$$
I:(X_1^*,\|\cdot\|_{1,*})\longrightarrow(X_2^*,\|\cdot\|_{2,*})
$$
has closed graph: if $f_n\to f$ in the first dual norm and $f_n\to g$ in the second, evaluating at each $x\in V$ gives $f(x)=g(x)$. The <closed graph theorem> makes $I$ bounded, and the same argument for $I^{-1}$ makes the dual norms equivalent. Thus there are $c,C>0$ with
$$
c\|f\|_{1,*}\leq\|f\|_{2,*}\leq C\|f\|_{1,*}.
$$
The dual formula
$$
\|x\|_j=\sup_{\|f\|_{j,*}\leq1}|f(x)|
$$
from the Hahn--Banach theorem transfers these inequalities to the original norms. Hence $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.
Solved by gpt-5.6-sol high.
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