Compact and Hilbert–Schmidt Operators

If $T$ is compact, then $T$ is bounded.

If $T$ is bounded and has finite rank, then $T$ is compact.

If $T_{n}: X \rightarrow Y$ are compact operators and $\left\| T-T_{n}\right\| \rightarrow 0$, then $T$ is compact.

If $T$ is compact and $S$ is bounded, or if $T$ is bounded and $S$ is compact, then $S T$ is compact.

If $T: X \rightarrow X$ is compact and $\lambda \in \mathbf{F}$, then $\ker (T-\lambda I)$ is finite dimensional.

Show that $M_{\lambda }$ is compact if and only if $\lambda \in c_{0}$.

Show that if $\lambda \in c_{0}$ and $\lambda_{n} \neq 0$ for every $n$, then $\left\{ M_{\lambda } x:\left\| x\right\| \leq 1\right\}$ is not closed in $H$.

Given $T \in \mathcal{B}(H, K)$, the quantity $\left\| T\right\|_{\mathrm{HS}}^{2} = \sum_{n}\left\| T e_{n}\right\|^{2}$ is independent of the choice of orthonormal basis $\left\{ e_{n}\right\}$ for $H$ (i.e., it is either finite and equal for all orthonormal bases, or infinite for every orthonormal basis). Further, $\left\| T^{*}\right\|_{\mathrm{HS}} = \left\| T\right\|_{\mathrm{HS}}$ and $\left\| T\right\| \leq \left\| T\right\|_{\mathrm{HS}}$.

$\left\| \cdot \right\|_{\mathrm{HS}}$ is a norm on $\mathcal{B}_{2}(H, K)$, and $\mathcal{B}_{2}(H, K)$ is complete with respect to this norm. Further, $\mathcal{B}_{2}(H, K)$ is a Hilbert space with respect to the inner product $\left\langle T, U \right\rangle_{\mathrm{HS}} = \sum_{n} \left\langle T e_{n}, U e_{n} \right\rangle$, $T, U \in \mathcal{B}_{2}(H, K)$, where $\left\{ e_{n}\right\}$ is any orthonormal basis for $H$.

If $T \in \mathcal{B}_{2}(H, K)$, $A \in \mathcal{B}(H)$, and $B \in \mathcal{B}(K)$, then $B T, T A \in \mathcal{B}_{2}(H, K)$, and we have $\left\| B T\right\|_{\mathrm{HS}} \leq \left\| B\right\| \left\| T\right\|_{\mathrm{HS}}$, $\left\| T A\right\|_{\mathrm{HS}} \leq \left\| A\right\| \left\| T\right\|_{\mathrm{HS}}$.

All bounded linear finite-rank operators on $H$ are Hilbert-Schmidt, and all Hilbert-Schmidt operators are compact. Moreover, $\mathcal{B}_{00}(H, K) = \left\{ L \in \mathcal{B}(H, K): L \text{ has finite rank}\right\}$ is dense in $\mathcal{B}_{2}(H, K)$ with respect to $\left\| \cdot \right\|_{\mathrm{HS}}$.

$T \in \mathcal{B}(H, K)$ has rank one if and only if $T = x \otimes y$ for some $x \in H, y \in K$ not both zero.

$T \in \mathcal{B}(H, K)$ has finite rank if and only if $T = \sum_{k = 1}^{N} x_{k} \otimes y_{k}$ for some $N \in \mathbb {N}$, $x_{k} \in H$, and $y_{k} \in K$.

$\left\{ x \otimes y: x \in H, y \in K\right\}$ is complete in $\mathcal{B}_{2}(H, K)$.

If $x_{1}, x_{2} \in H$ and $y_{1}, y_{2} \in K$, then $\left\langle x_{1} \otimes y_{1}, x_{2} \otimes y_{2} \right\rangle_{\mathrm{HS}} = \left\langle x_{2}, x_{1} \right\rangle_{H} \left\langle y_{1}, y_{2} \right\rangle_{K}$.

If $\left\{ e_{n}\right\}$ is an orthonormal basis for $H$ and $\left\{ f_{n}\right\}$ is an orthonormal basis for $K$, then $\left\{ e_{m} \otimes f_{n}\right\}_{m, n \in \mathbb {N}}$ is an orthonormal basis for $\mathcal{B}_{2}(H, K)$.

If $T \in \mathcal{B}(H, K)$, then $\left\langle T, e_{m} \otimes f_{n} \right\rangle = \left\langle T e_{m}, f_{n} \right\rangle$ for every $m, n \in \mathbb {N}$.

$\left\| x \otimes y\right\|_{\mathrm{HS}} = \left\| x\right\|_{H}\left\| y\right\|_{K}$.

$(a x) \otimes y = \bar{a}(x \otimes y)$ and $x \otimes (b y) = b(x \otimes y)$.

$(x+w) \otimes y = (x \otimes y)+(w \otimes y)$ and $x \otimes (y+z) = (x \otimes y)+(x \otimes z)$.

$\left\| x \otimes y-w \otimes z\right\|_{\mathrm{HS}} \leq \left\| x-w\right\|_{H}\left\| y\right\|_{K}+\left\| w\right\|_{H}\left\| y-z\right\|_{K}$.

Given $x_{k} \in H$ and $y_{k} \in K$, show that the adjoint of $T = \sum_{k = 1}^{M} x_{k} \otimes y_{k}$ is $T^{*} = \sum_{k = 1}^{M} y_{k} \otimes x_{k}$

Given $T \in \mathcal{B}(H, K)$, show that $T$ is compact if and only if $T^{*}$ is compact.

If $x \in H$ and $y \in K$ then $V(x \otimes y) U^{*} = U x \otimes V y$.

There exists a unique bounded operator $T \in \mathcal{B}(H \otimes K)$ such that $\left\| T\right\| \leq \left\| U\right\| \left\| V\right\|$ and $T(g \otimes h) = T g \otimes T h$ for all $g \in H$ and $h \in K$. We call $T$ the tensor product of $U$ and $V$, and write $T = U \otimes V$.

$\left(U_{1} \otimes V_{1}\right)\left(U_{2} \otimes V_{2}\right) = U_{1} U_{2} \otimes V_{1} V_{2}$.

If $U$ and $V$ are each topological isomorphisms, then so is $T = U \otimes V$.

Specialize to $H = L^{2}(E)$ and $K = L^{2}(F)$. In particular, show that if $U \in \mathcal{B}\left(L^{2}(E)\right)$ and $V \in \mathcal{B}\left(L^{2}(R)\right)$ then there exists a unique operator $U \otimes V \in \mathcal{B}\left(L^{2}(E \times F)\right)$ such that $(U \otimes V)(g \otimes h) = U g \otimes V h$, where $g \otimes h$ and $U g \otimes V h$ are the functions defined by equation (B.3).