Past exam of the functional analysis II course of the University of Oxford / 2023 / 2 / b / iii

For $n\ge 1$ put $C_n=\overline{T(n{\overline{B}}_X)}$, with closure taken in $Y$. ThenIf every $C_n$ has empty interior, then each $C_n$ is a nowhere dense set, because it is closed. Each $T(n{\overline{B}}_X)$ is then a nowhere dense set too. Consequently $T(X)$ is a meagre set, that is, a set of the first category in $Y$.

Otherwise some $C_n$ has nonempty interior. Scaling gives nonempty interior of $\overline{T({\overline{B}}_X)}$. The geometric correction for approximate surjectivity proved in the preceding part gives an open ball about zero inside $T({\overline{B}}_X)$. Since $T(X)$ is a vector subspace, it then equals $Y$: scale any $y\in Y$ into this ball and scale its preimage back. Hence $T(X)$ is a closed set.

Thus, in fact, the stronger operator range category dichotomy holds:In particular $T(X)$ is closed or of the first category, as required. The only substantive theorem used here is the preceding geometric correction for approximate surjectivity: for a bounded linear operator from a Banach space to a normed space, nonempty interior of the closure of the image of the unit ball implies nonempty interior of the image itself. Its series proof was given there, so this proof does not require the Baire category theorem or completeness of $Y$.