Riesz's Lemma/Proof 2

Theorem

Let $X$ be a normed vector space.

Let $Y$ be a proper closed linear subspace of $X$.

Let $\alpha \in \openint 0 1$.

Then there exists $x_\alpha \in X$ such that:

$\norm {x_\alpha} = 1$

with:

$\norm {x_\alpha - y} > \alpha$

for all $y \in Y$.

Proof

Consider the normed quotient vector space $X / Y$ with quotient mapping $\pi$.

From Operator Norm of Quotient Mapping in Quotient Normed Vector Space is 1, we have:

$\norm \pi_{\map B {X, X/Y} } = 1$

Since $\alpha \in \openint 0 1$, there exists $x_\alpha \in X$ with $\norm {x_\alpha} = 1$ and:

$\norm {\map \pi {x_\alpha} }_{X/Y} > \alpha$

by the definition of the norm on the space of bounded linear transformations.

That is, by the definition of the quotient norm:

$\ds \inf_{z \mathop \in Y} \norm {x_\alpha - z} > \alpha$

So there exists $y \in Y$ such that:

$\norm {x_\alpha - y} > \alpha$

$\blacksquare$