Compact Operators on Banach Spaces

Let $X, Y$ be Banach spaces and let $K \in B(X,Y)$. Call $K$ compact if $\overline{K(\text{ball }X)}$ is compact in $Y$.

Definition 6.0.1   Call $T \in B(X,Y)$ completely continuous if for all sequences $(x_n) \in X$ with $x_n \rightarrow x$ weakly, $T x_n \rightarrow T x$ in norm.

Proposition 6.0.2   For $T \in B(X,Y)$,

1. $T$ compact $\implies$ $T$ completely continuous.
2. If $X$ is reflexive, $T$ completely continuous, $T$ is compact.

Example: If $T \in B(\ell^1, \ell^1)$, then $T$ is completely continuous (by HW). So the identity operator on $\ell^1$ is not compact.

Proof.

1. Suppose $x_n \rightarrow 0$ weakly. By PUB, $M = \sup \Vert x_n\Vert < \infty$. WLOG, $M \leq 1$ and so
   $$(T x_n ) \subset \overline{T(\text{ball }X)}$$
   By compactness, $\exists (x_{n_k}), y \in \overline{T(\text{ball }X)}$ such that $T x_{n_k} \rightarrow y$ in norm. But $T x_{n_k} \rightarrow y$ weakly; but $T x_{n_k} \rightarrow 0$ by the definition, so $y = 0$. $T$ is completely continuous.
2. Assume $X$ is reflexive and separable; $($ball $X,$   wk $^*)$ is a compact metric space (compact by reflexive, metric by separable). Let $(x_n)$ be a sequence in ball $X$. There is $(x_{n_k})$ such that $x_{n_k} \rightarrow x$ weakly for some $x \in$   ball $X$. By complete continuity of $T$, $T x_{n_k} \rightarrow T x$ in norm. $\overline{T(\text{ball }X)}$ is sequentially compact, i.e., compact.
   

   In general, let $(x_n)$ be a sequence in $X$. Let $Z = \overline{\text{span }}\{x_n, n \in \mathbb{N}\}$. Then, $Z \leq X$ and $Z$ is separable and reflexive. If $U = T\vert _Z$, then $U$ is also completely continuous, and so, by the special case, $U$ is compact. But $T x_n = U x_n$ for all $n \in \mathbb{N}$. So, $(T x_n)$ has a convergent subsequence. Hence, $T$ is compact.

$\qedsymbol$

Theorem 6.0.3   (Schauder). For $T \in B(X,Y)$, $T$ is compact iff $T^*$ is compact.

Proof. Let $(f_n)$ be a sequence in ball $Y^*$. By Alaoglu, $f_n$ has a wk $^*$-convergent subsequence $f_{n_k}$ with $f_{n_k} \rightarrow g$ for some $g \in$   ball $Y^*$.

Claim: $T^* f_n \rightarrow T^* g$. Let $\epsilon > 0$, $N \geq 1$. Pick $\frac\epsilon 3$-net for $\overline{T(\text{ball }X)}$, i.e., $y_1, \ldots, y_m$ such that $\overline{T(\text{ball }X)} \subset \bigcup_{j=1}^m B_{y_j} (\frac\epsilon 3)$. As $f_n \rightarrow g$ wk $^*$, there is $M \geq N$ such that

$$\vert(f_n - g)(y_\ell)\vert < \frac\epsilon 3, i = 1, \ldots, m$$

Let $x \in$   ball $X$ and pick $y_k$ such that $\Vert Tx - y_k \Vert < \frac\epsilon 3$. Then,

$$(T^* f_n - T^* g)(x) = (f_n - g)(Tx)$$

$$\leq \vert(f_n - g)(Tx - y_k)\vert + \vert(f_n - g)(y_k)\vert \leq 2\Vert Tx - y_k\Vert + \frac\epsilon 3 = \epsilon$$

Thus, $\Vert T^* f_n - T^* g\Vert \leq \epsilon$ for $n \geq M$. So, $T^*$ is compact.

For the reverse direction, $T^{**} : X^{**} \rightarrow Y^{**}$ is compact. Notice $T^{**}\vert _X$ is also compact. But, by an ``obvious" property, $T^{**}\vert _X = T$. So, $T$ is compact.
$\qedsymbol$

Let $B_0(X,Y)$ denote the compact operators. Let $B_{00}(X,Y)$ be the finite rank operators. Imitating the proofs for operators on Hilbert spaces,

1. $B_0(X,Y) \leq B(X,Y)$.
2. the product of a compact operator and a bounded operator is compact.
3. $B_0(X)$ is a closed two-sided ideal in $B(X)$.
   Definition 6.0.4   We say that $(x_n)_{n=1}^\infty$ in $X$ is a Schauder basis if for every $x \in X$, there is a unique sequence of scalars $(\alpha_n)$ such that $X = \sum_{n=1}^\infty \alpha_n x_n$ (converges in norm).

4. If $X$ has a Schauder basis, then $B_{00}(X)$ is (norm) )dense in $B_0(X)$.
   Counterexample: Enflo, in 1973, gave a reflexive separable Banach space, $X$, such that $B_{00}(X)$ is not dense in $B_0(X)$. In particular, $X$ does not have a Schauder basis. On the other hand, all the ``classical" Banach spaces have the finite ranks norm-dense in the compacts.
   

Sample Result: $B_{00}(C(X))$ is norm-dense in $B_0(C(X))$ for $X$ compact.

Recall:

1. $\mathcal{F}\subset C(X)$ is compact iff $\mathcal{F}$ is closed, bounded, and equicontinuous.
2. For a finite open cover $\mathcal{C}$ of a topological space $X$, a partition of unity subordinate to $\mathcal{C}$ is a set of continuous $f_C : X \rightarrow [0,1]$, $c \in \mathcal{C}$
   $$\sum _{c \in \mathcal{C}} f_c \equiv 1$$
   $$f_c(x) = 0, \forall x \notin c$$
   If $X$ is normal, then there is a partition of unity subordinate to each finite open cover.
   Proof. Let $T \in B_0(C(X)), \epsilon > 0$. Goal: Build $F \in B_{00}(C(X))$ such that $\Vert T - F\Vert < \epsilon$. By Arzela-Ascola, as $\overline{T(\text{ball }C(X))}$ is compact, it is equicontinuous. Thus, for each $x \in X$, there is an open set $U_x$ with $x \in U_x$,

   $$\vert(Tf)(y) - (Tf)(x)\vert < \epsilon$$

   for all $y \in U_x$, $f \in$   ball $C(X)$. By compactness, there is $x_1, \ldots, x_n$ such that $X \subset \bigcup_{i=1}^n U_{x_i}$. There is a partition of unity $f_1, \ldots, f_n$ subordinate to $U_{x_1}, \ldots, U_{x_n}$.
   

   Define $F:C(X) \rightarrow C(Y)$ by

   $$F f = \sum_{i=1}^n [ (Tf)(x_i)] f_i$$

   Clearly, $F \in B_{00}(C(X))$ as $\textrm{ran }F \subset V\{f_1, \ldots, f_n\}$.
   

   Claim: $\Vert T - F\Vert \leq \epsilon$:

   $$\vert(Tf)(x) - Ff(x)\vert = \vert\sum_{j=1}^n (Tf)(x) - f_j(x) - \sum_{j=1}^n (Tf)(x_j) f_j(x)\vert$$

   $$= \vert\sum_{j=1}^n [ Tf(x) - Tf(x_j)] f_j(x) \vert \leq \sum_{j=1}^n \vert Tf(x) - Tf(x)\vert f_j(x)$$

   $$< \epsilon$$

   $$f_j(x) \geq 0 \implies x \in U_x \implies \vert(Tf)(x) - (Tf)(x_j)\vert < \epsilon$$

   $\qedsymbol$

Friday, March 31, 2006: