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Spring 2016 - Problem 4

Banach spaces, Hahn-Banach

Let $V_1$ be a finite-dimensional subspace of the Banach space $V$ . Show that there exists a continuous projection $\func{P}{V}{V_1}$ , i.e., a continuous linear map $\func{P}{V}{V}$ such that $P^2 = P$ and the range of $P$ is equal to $V_1$ .

Solution.

Let $\set{x_1, \ldots, x_n}$ be a basis for $V_1$ . Normalizing, we may assume that $\norm{x_i} = 1$ for each $i$ . We have a canonical isomorphism $\func{\phi}{V_1}{\R^n}$ , via

\sum_{i=1}^n c_ix_i \mapsto \sum_{i=1}^n c_ie_i

In $\R^n$ , we have the projection map $\func{\pi_i}{\R^n}{\R}$ onto the $i$ -th component, and so $\func{\alpha_i}{V_1}{\R}$ given by $\alpha_i = \pi_i \circ \phi$ is a bounded linear functional on $V_1$ . By Hahn-Banach, they all admit extensions to $\alpha_i \in V^*$ , so we define

P\p{x} = \sum_{i=1}^n \alpha_i\p{x} x_i.

$P$ is linear by construction and is continuous since

\norm{P\p{x} - P\p{y}} \leq \p{\sum_{i=1}^n \norm{\alpha_i}}\norm{x - y},

and for any $x \in V_1$ ,

P\p{x} = \sum_{i=1}^n \alpha_i\p{x}x_i = x

by construction. Thus, $P^2 = P$ and $\im{P} = V_1$ , which completes the proof.