Banach Lecture 14

This lecture ( lecture14) was given on 9/28/09

Excerpts

Let $K$ be a compact metric space and $J \subset K$ closed.  The ideal of $J$ is the $I(J) = \{f \in C(K)| f|J = 0\}$.

Thm: $I(J)$ is complemeted in $C(K)$.

Then two examples were presented which show the converse of the corollary from last lecture doesn’t hold.

Another example, for $1 \le p \le \infty$, $H^p = \{f \in L^p(\mathbb{T})| \hat f(n) = 0, n = -1, -2, ...\}$. Recall $\mathbb{T} = S^1 \subset \mathbb{C}$.  It turns out this subspace is complemented in $L^p(\mathbb{T})$ for $1 < p < \infty$ and not complemented for $p = 1, \infty$.

Compact Operators

$K \in \mathscr{L}(X,Y)$, for $X,Y$ Banach spaces, is compact if it maps bounded sets to relatively bounded sets (image has compact closure).

1. if $K$ is linear and maps bounded sets to relatively compact sets, then $K$ is bounded
2. if $K$ maps the unit ball in $X$ to a relatively bounded set in $Y$, then $K$ is compact.
3. $K$ is compact iff for every bounded sequence $(x_n)_1^\infty \subset X$, the sequence $(Kx_n)_1^\infty \subset Y$ has a convergent subsequence.
4. operators of finite rank are compact
5. the identity operator on $X$ is compact iff $X$ is finite dimensional.
6. The sum of two compact operators is compact.
7. The product of a compact operator with a bounded operator (in either order) is compact; this implies an idempotent is compact iff its image has finite rank.

Prop. The set of compact operators in $\mathscr{L}(X,Y)$ is closed in the operator norm.