Further Properties

So for we have proved that $K(H)$ is closed subspace and closed ideal of $B(H)$.

Exercise 2   If $F\in B(H)$ is a finite rank operator then

1. there exists $\phi_1, \dots \phi_n$ and $\psi_1, \dots \psi_n$ in $H$ such that $Fh=\sum_{k=1}^{n}\langle h,\phi_k\rangle \psi_k$
2. $F^*$ also a finite rank operator.

Theorem 5.1   If $T\in K(H)$, then $\overline{R(T)}$ is separable.

Theorem 5.2   If $T\in K(H)$, then there exists a sequence of finite rank operators $F_n$ such that $\Vert F_n-T\Vert\rightarrow 0$. In fact $F_n=P_nT$

Theorem 5.3   Let $T\in B(H)$. $T$ is compact $\Leftrightarrow$ $T^*$ is compact

Theorem 5.4   Let $T\in B(H)$.

$$T\in K(H)\Leftrightarrow T^*\in K(H)\Leftrightarrow T^*T\in K(H)\Leftrightarrow TT^*$$

Exercise 3   Let $e_n$ be a ONB of a separable Hilbert space $H$. Let $\{\lambda_n\}_{n\in \mathbb{N}}$ be a bounded sequence. Defeine $A: H\rightarrow H$ as

$$Ah=\sum_{n=1}^{\infty}\lambda_n\langle h,e_n\rangle e_n$$    and $$A_Nh=\sum_{n=1}^{N}\lambda_n\langle h,e_n\rangle e_n$$

Then

$$\Vert(A-A_N)h\Vert^2\leq \sup_{n>N} \vert\lambda_n\vert^2\Vert h\Vert^2=\limsup_{n\rightarrow \infty} \vert\lambda_n\vert^2$$

Definition 4 (eigenvalue, eigen vector, point spectrum $\sigma_p(T)$)  

Theorem 5.5   If $T\in K(H)$ and $\lambda \in \sigma_p(T)$ and $\lambda\neq 0$ then $N(T-\lambda I)$ is finite dimensional.

Definition 5   $T\in B(H)$ is said to be bounded below if there exists $m>0$ such that $\Vert Th\Vert\geq m \Vert h\Vert$.

Note that if $T$ is bounded below then $T$ is one-one. In special case if $T-\lambda$ is bounded below then $T-\lambda$ is one-one. Converse is not true in general but when $T$ is compact and $\lambda\neq 0$ then the converse holds.

Exercise 4   If $T\in B(H)$ is bounded below then $R(T)$ is closed. (Hint: Cauchy sequence)

Theorem 5.6   Let $T\in K(H)$ and $\lambda\neq 0$. If $\inf_{\Vert h\Vert=1}\Vert Th-\lambda h\Vert=0$ then $\lambda \in \sigma_p(T)$.

Corollary 1   Let $T\in K(H)$ and $0 \neq \lambda \notin \sigma_p(T)$ then $\inf_{\Vert h\Vert=1}\Vert Th-\lambda h\Vert>0$. That is $T-\lambda$ is bounded below.

Corollary 2   Suppose $T\in K(H)$ and $0 \neq \lambda \notin \sigma_p(T)$ and $\overline{\lambda}\notin \sigma_p(T^*)$ , then $T-\lambda$ is a bijection and $(T-\lambda I)^{-1}$ is a bounded operator.

Theorem 5.7   If $T\in K(H)$ and $T=T^*$ then either $\Vert T\Vert$ or $-\Vert T\Vert$ is an eigen value of $T$.

Theorem 5.8 (Spectral Theorem- Compact Self-Adjoint)   Let $T\in K(H)$ and $T=T^*$ then it has only countable number of eigen value $\{\lambda_n\}$ and ONB $\{\phi_n\}$ for $\overline{R(T)}$ such that

$$Th=\sum_{n=1}^{\infty}\lambda_n \langle h,\phi_n\rangle \phi_n= \sum_{n=1}^{\infty}\lambda_n P_nh$$

1. $\vert\lambda_1\vert\geq \vert\lambda_2\vert\geq \dots$ If $\{\lambda_n\}$ are countably infinite then $\{\lambda_n\}\rightarrow 0$.
2. $\{\phi_n\}$ are eigen vectors corresponding to eigen value $\{\lambda_n\}$
3. $N(T-\lambda_n)$ is finite dimensional for each $n$.
4. $N(T-\lambda_n)\perp N(T-\lambda_m)$ when $\lambda_n\neq \lambda_m$

Proof.

1. $\overline{R(T)}$ seperable.
2. $\Vert T\Vert$ or $-\Vert T\Vert$ is an eigen value of $T$
3. The

$\qedsymbol$