Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 22H (Linear Analysis)

For a complex Banach space, the spectrum of a bounded operator and resolvent set areThe bounded inverse theorem says that a bounded bijection between Banach spaces automatically has a bounded inverse, so failure of bijectivity suffices here. The point spectrum isthe set of eigenvalues. For a real Banach space, these definitions use its complexification.

A compact operator is a bounded linear operator carrying every bounded set into a relatively compact set. Equivalently, for every bounded sequence $(x_n)$, the sequence $(T{x}_n)$ has a norm-convergent subsequence. It is enough to require this for the closed unit ball, because any bounded set lies in a scalar multiple of that ball.

Suppose otherwise and choose linearly independent eigenvectors $v_n$ with $T{v}_n={\lambda }_n{v}_n$ and $|{\lambda }_n|>a$. Put $F_n=\mathrm{span}(v_1,\dots ,v_n)$. Each $F_n$ is finite-dimensional and $T$-invariant. Moreover $(T-{\lambda }_{n}I)F_n\subseteq F_{n-1}$, including when some eigenvalues coincide.

Apply the supplied distance fact inside the Banach space $F_n$ to choose $x_n\in F_n$ with $\Vert x_n\Vert =1$ and $\mathrm{dist}(x_n,F_{n-1})=1$. For $m<n$, both $(T-{\lambda }_{n}I)x_n$ and $T{x}_m$ belong to $F_{n-1}$. ConsequentlyThus $(T{x}_n)$ has no Cauchy subsequence, although $(x_n)$ is bounded. This contradicts the compact operator property. Only finitely many linearly independent eigenvectors can have eigenvalue modulus greater than $a$.

The exact condition for this diagonal operator on sequence space isFirst, $T$ is bounded precisely when ${\sup }_n|{\lambda }_n|<\infty$, with operator norm equal to that supremum: the upper bound follows by summing $|{\lambda }_n{x}_n{|}^2$, and the coordinate unit vectors $e_n$ give the lower bound. If the supremum is infinite, choose distinct indices with $|{\lambda }_{n_j}|\ge j$ and set $x_{n_j}=1/j$, other coordinates zero. Then $x\in {\ell }^2$ but $Tx\notin {\ell }^2$, so the displayed prescription would not even define an everywhere-defined map into ${\ell }^2$.

If $T$ is compact but ${\lambda }_n\nrightarrow 0$, infinitely many indices satisfy $|{\lambda }_n|\ge ϵ>0$. Along those indices,contradicting compactness. Conversely, if ${\lambda }_n\to 0$, truncate to $T_{N}x=({\lambda }_1{x}_1,\dots ,{\lambda }_N{x}_N,0,\dots )$. This is a finite-rank compact operator, and $\Vert T-T_N\Vert ={\sup }_{n>N}|{\lambda }_n|\to 0$. To see compactness directly, approximate the image of the unit ball by $T_N$ to within $ϵ/2$, and cover its bounded finite-dimensional image by finitely many $ϵ/2$-balls. Hence $T$'s unit-ball image is totally bounded; its closure is compact because ${\ell }^2$ is complete.