Answer to Question #164918 in Functional Analysis for Ebenezer

Question

Let V be a Banach space and let M ⊆ V be a proper subspace of V (i.e. M not equal to V ). Prove that if v ∈ V and v /∈ M then there is a φ ∈ V* such that φ(v) = 1 and φ(w) = 0 for every w ∈ M.

Answer 1

First of all, we need to have $\bar M \neq V, v\notin \bar M$, otherwise it would not be possible to construct such a form.

Define $N=M\oplus \mathbb{K}v$ and now for any $w \in N, w=\lambda_M\cdot m+\lambda_v\cdot v$, where $m\in M$. Now we can define a linear form $\phi_N :(\lambda_M \cdot m + \lambda_v \cdot v)\mapsto \lambda_v$. Let's prove that it is continuous by calculating it's norm $||\phi_N|| = \sup_{w\in N}\frac{|\lambda_v |}{||\lambda_M\cdot m + \lambda_v \cdot v||}$, $||\phi_N|| =\sup_{w\in N} \frac{1}{||v+\frac{\lambda_M \cdot m}{|\lambda_v|}||}$ , the vector $\frac{\lambda_M}{|\lambda_v|}m$ is in $M$ and as $v\notin \bar M$, the expression below is bounded below by some constant (as otherwise the vector $v$ could be approached by $-\frac{\lambda_M}{|\lambda_v|}m$ as closely, as we want and thus $v\in \bar M$). Therefore the norm $||\phi_N||$ is bounded, so $\phi_N$ is continuous. Now by Hahn-Banach's theorem we can extend $\phi_N$ on $V$ such that the extension $\phi|_M=0, \phi(v)=1, \phi$ is continuous.