8

A Banach space $X$ is said to be injective if for all Banach spaces $W,Z$ with $W\subset Z$, and operators $T\in B(W,X)$, $T$ can be extended to all of $Z$ with the same norm.

Equivalently, $X$ is injective if it is complemented by a norm $1$ projection in any Banach space containing it.

Labelling the first definition as $(1)$ and the second definition as $(2)$, the proof for $(1)\Rightarrow (2)$ is brief.

If $X\subset Y$ for some Banach space $Y$, then applying $(1)$ to the identity map on $X$ yields the projection.

The $(2)\Rightarrow (1)$ direction I have been stuck on for a couple of days. Can anyone offer a hint? Thanks very much in advance!

roo
  • 5,598

1 Answers1

10

You can isometrically embed $X$ into $\ell^\infty(\Gamma)$ where $\Gamma$ is some humongous set of indices (e.g., take every point of the unit ball of $X^*$ as an index).

Given $T:W\to X$, think of it as $T:W\to \ell^\infty(\Gamma)$ and extend it to $\widetilde T: Z\to \ell^\infty(\Gamma)$ preserving the norm (the special form of the norm on $\ell^\infty(\Gamma)$ will help.) Then do the obvious thing.

user
  • 1,012
  • Thanks! I will give this a shot. I kept trying to embed $X$ into some sort of direct sum, but it never went anywhere. – roo Aug 19 '13 at 03:37
  • Got it, just take $P\circ\tilde{T}$ as the extension, where $P$ is the projection which exists by assumption. Thanks!

    Yes after typing our my first response I immediately realized that it made no sense.

     – roo Aug 19 '13 at 04:23