Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Lecture 42: Bases of eigenvectors for compact self adjoint operators

Consider the sequence $e_1,e_2,e_3,\dots$ in ${\ell }^p$ where $$e_i=(0,0,\dots ,0,1,0,0,\dots )$$ with $1$ in the $i\text{th}$ spot. Then

(a)	$e_1,e_2,e_3,\dots$ has no convergent subsequence. (The closed unit ball is not compact.)
(b)	$e_1,e_2,e_3$ weakly converges to $0$ i.e. If $x\in {\ell }^2,$ $x=(x_1,x_2,\dots ),$ then $⟨x,e_1⟩,⟨x,e_2⟩,⟨x,e_3⟩$ converges to $0\text{.}$
(c)	Let $I:{\ell }^p\to {\ell }^p$ be the identity operator. Then $I{e}_1,I{e}_2,I{e}_3,\dots$ is the sequence $e_1,e_2,e_3,\dots$ has no convergent subsequence and so $\overline{I\left(S\right)}$ is not compact.

Let $T:H\to H$ be a compact linear operator. Let $\lambda$ be a nonzero eigenvalue of $T$ and let $$X_{\lambda }=\{x\in H\hspace{0.17em}|\hspace{0.17em}Tx=\lambda x\}\text{.}$$ If $\text{dim}\left(X_{\lambda }\right)$ is infinite then there is a sequence $e_1,e_2,e_3,\dots$ in $X_{\lambda }$ of linearly independent vectors with $\Vert e_j\Vert =1$ and $T{e}_j=\lambda e_j$ and $⟨e_i,e_j⟩={\delta }_{ij}\text{.}$ Then $$\begin{array}{rcl}{\displaystyle {\Vert T{e}_m-T{e}_n\Vert }^2}& =& {\displaystyle {\Vert \lambda e_m-\lambda e_n\Vert }^2}\\ & =& {\displaystyle {\mid \lambda \mid }^2{\Vert e_m-e_n\Vert }^2}\\ & =& {\displaystyle {\mid \lambda \mid }^2({\Vert e_m\Vert }^2+{\Vert e_n\Vert }^2)}\\ & =& {\displaystyle {\mid \lambda \mid }^2·2\text{.}}\end{array}$$ So the sequence $T{e}_1,T{e}_2,\dots$ has no Cauchy subsequence and no convergent subsequence. So $T$ is not compact.

Let $T:H\to H$ be a compact linear operator and let ${\lambda }_1,{\lambda }_2,{\lambda }_3,\dots$ be distinct eigenvalues of $T\text{.}$
Assume $\underset{k\to \infty }{\text{lim}}{\lambda }_k\ne 0\text{.}$
Then there is a subsequence ${\lambda }_{k_1},{\lambda }_{k_2},\dots$ and $c\in {ℝ}_{>0}$ with $\mid {\lambda }_{k_j}\mid >C$ for $j\in {\mathbb{Z}}_{>0}\text{.}$
Let $e_1,e_2,\dots$ be such that $\Vert e_i\Vert =1$ and $T{e}_j={\lambda }_{k_j}{e}_j\text{.}$
Since ${\lambda }_{k_1},{\lambda }_{k_2},\dots$ are all distinct then $⟨e_i,e_j⟩=0\text{.}$
So $$\begin{array}{rcl}{\displaystyle {\Vert T{e}_m-T{e}_n\Vert }^2}& =& {\displaystyle {\Vert {\lambda }_m{e}_m-{\lambda }_n{e}_n\Vert }^2}\\ & =& {\displaystyle ⟨{\lambda }_m{e}_m-{\lambda }_n{e}_n,{\lambda }_m{e}_m-{\lambda }_n{e}_n⟩}\\ & =& {\displaystyle {\lambda }_m^2{\Vert e_m\Vert }^2+{\lambda }_n^2{\Vert e_n\Vert }^2}\\ & >& {\displaystyle 2C^2\text{.}}\end{array}$$ So $T{e}_1,T{e}_2,T{e}_3,\dots$ has no Cauchy subsequence and no convergent subsequence.
So $T$ is not compact.

Remark Let $H$ be a Hilbert space and let $T:H\to H$ be a compact operator. If $H$ is infinite dimensional then $T$ is not a bijection (or, better, $0·I-T$ is not a bijection).

Let $T:H\to H$ be a compact self adjoint operator. For each eigenvalue let $B_{\lambda }$ be an orthonormal basis of $X_{\lambda }$ and let $$B=\bigcup _{\text{eigenvalues}}{B}_{\lambda }$$ and let $$X=\overline{\text{span}\left(B\right)}\text{.}$$ Then $$H=X\oplus X^{\perp }\text{and}T=T_X\oplus T_{X^{\perp }}$$ with $T_X$ and $T_{X^{\perp }}$ both compact operators $$T_{X^{\perp }}:X^{\perp }⟶X^{\perp }\text{.}$$ Then $T_{X^{\perp }}$ has an eigenvector with eigenvalue $\Vert T_{X^{\perp }}\Vert \text{.}$ But all eigenspaces of $T$ are in $X\text{.}$ So $H=X\text{.}$

Notes and References

These are a typed copy of Lecture 42 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on October 12, 2014.

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