Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2010 / A / 2

The Baire category theorem states that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets; equivalently, every countable intersection of open dense sets is dense.

Let $E_n=\mathrm{span}\{e_1,\dots ,e_n\}$ in the space of finitely supported sequences. Each $E_n$ is a closed linear subspace for every norm, since a finite-dimensional subspace is closed. Henceis a closed linear subspace of the Banach space $X$. Every finitely supported sequence lies in some $E_n$, so $X={\bigcup }_{n\ge 1}{F}_n$. By the Baire category theorem, some $F_N$ has nonempty interior.

A linear subspace with nonempty interior is the whole space: translate a ball in it to zero and scale any vector into that ball. Thus $F_N=X$, giving $T(X)\subseteq E_N$ andThe proof uses no particular choice of norm on the space of finitely supported sequences.

Under duality of sequence spaces, $a=(a_n)\in {\ell }_1$ represents $f_a(x)=\sum a_n{x}_n$ on $c_0$, with operator norm $\Vert f_a\Vert =\sum |a_n|$. Here $c_0$ is the space of sequences converging to zero, equipped with its supremum norm.

If $\Vert x{\Vert }_{\infty }\le 1$ and $f_a(x)=\sum |a_n|$, thenEvery summand is nonnegative, so $x_n=\mathrm{sgn}(a_n)$ whenever $a_n\ne 0$. Infinitely many such coordinates would give infinitely many $|x_n|=1$, contradicting $x_n\to 0$. Therefore $a$ has finite support.

Conversely, if $a$ has finite support, choose $x_n=\mathrm{sgn}(a_n)$ on that support and $x_n=0$ elsewhere. Then $x\in c_0$, $\Vert x{\Vert }_{\infty }\le 1$, and $f_a(x)=\Vert f_a\Vert$. The zero functional is included by taking $x=0$. ThusThis is norm attainment on the vanishing sequence space.

The compact set $S(B_{c_0})$ is bounded, so $S$ is a bounded linear operator. For each $g\in Y^{\ast }$, the continuous linear functional $g$ attains its maximum on that compact set. Symmetry of $S(B_{c_0})$ implies that maximum is $\Vert S^{\ast }g\Vert$, so $S^{\ast }g$ attains its norm on $B_{c_0}$. By norm attainment on the vanishing sequence space,Regard the Banach-space adjoint $S^{\ast }$ as a bounded linear operator from the Banach space $Y^{\ast }$ into $c_{00}$ with its inherited ${\ell }_1$ norm. The completeness of the dual space holds even if $Y$ is incomplete. The preceding Baire category theorem argument shows that $S^{\ast }$ has finite rank.

If $S^{\ast }(Y^{\ast })=\mathrm{span}\{f_1,\dots ,f_r\}$, set $N={\bigcap }_{j=1}^r\ker f_j$. For $x\in N$ and every $g\in Y^{\ast }$, $g(Sx)=(S^{\ast }g)(x)=0$. The Hahn-Banach theorem implies that continuous linear functionals separate points of $Y$, so $Sx=0$. Thus $N\subseteq \ker S$, andThis proves compact closed-ball image forces finite rank on c0.

Define a diagonal operator on sequence space byIt is bounded because $\Vert Sx{\Vert }_2^2\le \Vert x{\Vert }_{\infty }^2{\sum }_{n\ge 1}{4}^{-n}=\Vert x{\Vert }_{\infty }^2/3$. Its truncation $S_{m}x=(2^{-1}{x}_1,\dots ,2^{-m}{x}_m,0,\dots )$ is a finite-rank operator, andEquality follows by using finitely supported sequences equal to $1$ on longer and longer finite tails. An operator norm limit of compact operators into the Banach space ${\ell }_2$ is compact; therefore $S$ is a compact operator. The vectors $S{e}_n=2^{-n}{e}_n$ are linearly independent, so its rank is infinite.

There is no conflict with the preceding result about a compact closed-ball image. Indeed, take $x^{(m)}=(1,\dots ,1,0,\dots )$ with $m$ initial ones. Then $S{x}^{(m)}\to (2^{-n}{)}_{n\ge 1}$ in ${\ell }_2$, but that limit has no preimage in $c_0$: its only possible preimage is the constant sequence $1$. Thus $S(B_{c_0})$ is relatively compact but is not closed and hence is not compact.