# Why Hahn-Banach can not be generalized to operators

Let $E$ be a Banach Space, $F \subset E$ non-closed subspace, $T \in L(F,F)$ the identity. Then there is no $S \in L(E,F)$ extending $T$. Why?

• What could you say about $S(\overline{F})$ if an extension existed? Oct 14 '17 at 19:35
• Let $x\in\overline F\setminus F$. Then there exists $(x_n)\subset F$ with $x_n\to x$, and you have $x = \lim x_n = \lim Tx_n = \ldots$. Oct 14 '17 at 19:44
• Not quite sure what your point is
– Pazu
Oct 14 '17 at 21:11

The space $F$ is non-complete, so there are “not enough space” to place the continuous images of the points of its completion $\bar{F}\subset E$. The domain completeness (of some kind) is a typical condition in map extension theorems. The intuition here is similar to that we cannot extend the identity map $\operatorname{id}:\Bbb Q\to\Bbb Q$ from $\Bbb Q\subset \Bbb R$ to a continuous map from $\Bbb R$ to $\Bbb Q$.