Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 106 / 4 / c / 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 $X_j^{\ast }$ is a Banach space in its dual norm. The identityhas 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 $c,C>0$ withThe dual formulafrom the Hahn--Banach theorem transfers these inequalities to the original norms. Hence $\Vert \cdot {\Vert }_1$ and $\Vert \cdot {\Vert }_2$ are equivalent.

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