Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 21F (Linear Analysis)

Let $(T_m)$ be Cauchy in the operator norm on $\mathcal{B}(X,Y)$. For each $x\in X$,so $(T_{m}x)$ is Cauchy in the Banach space $Y$. DefinePointwise passage to the limit shows that $T$ is linear. A norm-Cauchy sequence is bounded, say $\Vert T_m\Vert \le M$, and henceThus $T$ is bounded. Finally, for fixed $m$,The right-hand coefficient tends to zero with $m$, so $T_m\to T$ in operator norm. ThereforeThis is the Banach space of bounded linear operators.

For $x\in X$, choose $x_m\in D$ with $x_m\to x$ and defineThe sequence on the right is Cauchy becauseand it converges because $Y$ is complete. The same estimate shows that the limit is independent of the approximating sequence, is linear, and satisfiesIts norm is at most $\Vert T\Vert$, while restriction to $D$ gives the reverse inequality, soAny two continuous extensions agree on the dense set $D$ and hence everywhere, proving uniqueness. This is the extension of a bounded linear operator from a dense subspace.

Completeness of $X$ was never used, so the result remains true when $X$ is incomplete. Completeness of $Y$ is essential. For example, with $D=c_{00}\subset X={\ell }^2$, let the codomain be the incomplete normed space $Y=c_{00}$ with the ${\ell }^2$ norm and let $T$ be the identity. An extension would have to send every limit in ${\ell }^2$ to itself and therefore could not take values in $c_{00}$. This is the failure of dense-subspace extension into an incomplete codomain.

For every finite scalar sequence $a=(a_n)$ with $|a_n|\le 1$, define a bounded linear functional on the continuous dual space $X^{\ast }$ byFor each fixed $f\in X^{\ast }$, the hypothesis giveswhere the supremum is over all finite choices and the phases $a_n$ may be chosen to align the complex numbers $f(x_n)$.

The dual $X^{\ast }$ is a Banach space even when $X$ is incomplete. The Uniform boundedness principle therefore gives a constant $C$ such thatFor each $N$, choose $a_n$ for $n\le N$ so that $a_{n}f(x_n)=|f(x_n)|$. ThenLetting $N\to \infty$ provesThis is the uniform bound from pointwise absolute summability of dual evaluations.