Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 22H (Linear Analysis)

If a subspace $W\subset V$ contains a ball about some $w\in W$, translation gives a ball about zero in $W$, and scaling then gives every vector of $V$. Thus every proper subspace has empty interior.

If a complete infinite-dimensional $V$ had a countable Hamel basis $(e_n)$, thenEach finite-dimensional subspace is closed and, being proper, nowhere dense. This contradicts the Baire category theorem. The polynomials have the countable Hamel basis $1,x,x^2,\dots$ and are infinite-dimensional, so no norm can make that whole vector space complete.

The identity mapis a continuous linear bijection. Since both spaces are Banach, the bounded inverse theorem makes $I^{-1}$ continuous, giving $\Vert x{\Vert }_1\le C_2\Vert x{\Vert }_2$.

Completeness is essential. On the space $c_{00}$ of finitely supported sequences, $\Vert x{\Vert }_2\le \Vert x{\Vert }_1$, but for $x=(1,\dots ,1,0,\dots )$ the ratio $\Vert x{\Vert }_1/\Vert x{\Vert }_2=\sqrt{n}$ is unbounded. Both normed spaces are incomplete.

DefineThe hypothesis makes $T$ everywhere defined. The dual $V^{\ast }$ is Banach even when $V$ is incomplete. If $f_j\to f$ in $V^{\ast }$ and $T{f}_j\to a$ in ${\ell }^1$, then coordinatewise convergence gives $a_n=f(x_n)$, so the graph of $T$ is closed. The closed graph theorem therefore givesTaking the supremum over the unit ball proves the required finite bound.