Compact Operators

Prove Proposition 4.2(c): If $A \in \mathcal{B}(\mathcal{H})$, $B \in \mathcal{B}(\mathcal{K})$, and $T \in \mathcal{B}_{0}(\mathcal{H}, \mathcal{K})$, then $TA$ and $BT \in \mathcal{B}_{0}(\mathcal{H}, \mathcal{K})$.

Show that every operator of finite rank is compact.

If $T \in \mathcal{B}_{00}(\mathcal{H}, \mathcal{K})$, show that $T^{*} \in \mathcal{B}_{00}(\mathcal{K}, \mathcal{H})$ and $\dim (\operatorname {ran} T) = \dim (\operatorname {ran} T^{*})$.

Show that an idempotent is compact if and only if it has finite rank.

Show that no nonzero multiplication operator on $L^{2}(0,1)$ is compact.

Show that if $T: \mathcal{H} \rightarrow \mathcal{K}$ is a compact operator and $\left\{ e_{n}\right\}$ is any orthonormal sequence in $\mathcal{H}$, then $\left\| Te_{n}\right\| \rightarrow 0$. Is the converse true?

If $T$ is compact and $\mathcal{M}$ is an invariant subspace for $T$, show that $T|\mathcal{M}$ is compact.

If $h, g \in \mathcal{K}$, define $T: \mathcal{H} \rightarrow \mathcal{K}$ by $Tf = \left\langle f, h \right\rangle g$. Show that $T$ has rank $1$ [that is, $\dim (\operatorname {ran} T) = 1$]. Moreover, every rank $1$ operator can be so represented. Show that if $T$ is a finite rank operator, then there are orthonormal vectors $e_{1}, \ldots , e_{n}$ and vectors $g_{1}, \ldots , g_{n}$ such that $Th = \sum_{j=1}^{n} \left\langle h, e_{j} \right\rangle g_{j}$ for all $h$ in $\mathcal{H}$. In this case show that $T$ is normal if $g_{j} = \lambda_{j} e_{j}$ for some scalars $\lambda_{1}, \ldots , \lambda_{n}$. Find $\sigma_{p}(T)$.

Show that a diagonalizable operator is normal.

Verify the statements in Example 4.10: Let $A$ be the diagonalizable operator in Proposition 4.6 (that is, $\mathcal{H}$ is a separable Hilbert space with basis $\left\{ e_{n}\right\}$; $\left\{ \alpha_{n}\right\} \subseteq \mathbb {F}$ with $M = \sup \left\{ \left|\alpha_{n}\right|: n \geq 1\right\} < \infty$; $Ae_{n} = \alpha_{n} e_{n}$ for all $n$). Then $\sigma_{p}(A) = \left\{ \alpha_{1}, \alpha_{2}, \ldots \right\}$. If $\alpha \in \sigma_{p}(A)$, let $J_{\alpha } = \left\{ j \in \mathbb {N}: \alpha_{j} = \alpha \right\}$. Then $h$ is an eigenvector for $\alpha$ if and only if $h \in \vee \left\{ e_{j}: j \in J_{\alpha }\right\}$.

Verify the statement in Example 4.11: The Volterra operator has no eigenvalues.

Verify the statement in Example 4.12: Let $h \in \mathcal{H} = L_{\mathbb {C}}^{2}(-\pi , \pi )$ and define $K: \mathcal{H} \rightarrow \mathcal{H}$ by $(Kf)(x) = (2\pi )^{-1/2} \int_{-\pi }^{\pi } h(x-y) f(y) \, dy$. If $\lambda_{n} = (2\pi )^{-1/2} \int_{-\pi }^{\pi } h(x) \exp (-inx) \, dx = \hat{h}(n)$, the $n$th Fourier coefficient of $h$, then $Ke_{n} = \lambda_{n} e_{n}$, where $e_{n}(x) = (2\pi )^{-1/2} \exp (-inx)$. (Note that the operator $K$ in this example is diagonalizable.)

If $T_{n} \in \mathcal{B}(\mathcal{H}_{n})$, $n \geq 1$, with $\sup_{n} \left\| T_{n}\right\| < \infty$ and $T = \oplus_{n=1}^{\infty } T_{n}$ on $\mathcal{H} = \oplus_{n=1}^{\infty } \mathcal{H}_{n}$, show that $T$ is compact if and only if each $T_{n}$ is compact and $\left\| T_{n}\right\| \rightarrow 0$.

In Lemma 4.8, show that if $L^{2}(X, \Omega , \mu )$ is separable, then $\left\{ \phi_{ij}\right\}$ is a basis for $L^{2}(X \times X, \Omega \times \Omega , \mu \times \mu )$. What if $L^{2}(X, \Omega , \mu )$ is not separable?

Lemma 4.8: If $\left\{ e_{i}: i \in I\right\}$ is a basis for $L^{2}(X, \Omega , \mu )$ and

for $i,j$ in $I$ and $x,y$ in $X$, then $\left\{ \phi_{ij}: i,j \in I\right\}$ is an orthonormal set in $L^{2}(X \times X, \Omega \times \Omega , \mu \times \mu )$. If $k$ and $K$ are as in Proposition 4.7, then $\left\langle k, \phi_{ij} \right\rangle = \left\langle Ke_{j}, e_{i} \right\rangle$.