Banach Lecture 14

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


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.


About this entry