Existence of a normalized vector, far from proper closed subspace.

Question

Say we have a normed vector space $(E, \|\cdot\|)$, and a proper closed subspace $F\subsetneq E$, I want to prove the following:

$$(\forall \varepsilon > 0), (\exists e \in E) : \|e\|=1 \text{ and } d(e,F) \geq 1 - \varepsilon $$

Where $d(e,F)= \inf\limits_{y \in F} \|e-y\|$.

I started proving the following fact:

Given $x \in E, \alpha \geq 1$, $\exists f \in F $ such that $\|x-f\|\leq\alpha \cdot d(x,F)$

Which is true by definition of $d$ as an infimum, because: $$ (\forall \varepsilon>0), (\exists f_{\varepsilon} \in F): d(x,F)\leq d(x,f_{\varepsilon})\leq d(x,F)+\varepsilon$$ And taking $\varepsilon = (\alpha - 1)\cdot d(x,F)$ we conclude.

So trying to use the last fact, let $\varepsilon >0,$ we define for $\bar{x} \notin F$ and $ \alpha \in \mathbb{R}_{+}$ to be determined, the $f\in F$ such that $ \|\bar{x}-f\|\leq \alpha\cdot d(\bar{x},F)$, then I don't know how to follow.

Mostly I don't know how to link $\bar{x}-f$ with the $e\in E $ that I'm trying to find, I could normalize, but I can't figure out where to get the inequality I'm looking for, my guess would be through fixing $\alpha = \frac{1}{1-\varepsilon}$ or something like that, but inequalities I manage to get are upper bounds.

Any help would be appreciated.

Answer 1

I managed to follow trough with David's help.

So we choose $y=\frac{\bar{x}-f}{\|\bar{x}-f\|}$, where f is given as in the property stated in the question body, for $\alpha =\frac{1}{1-\varepsilon}$.

Clearly $\|y\|=1$, so we compute it's distance to $F$.

$$d(y,F) = \inf\limits_{g \in F} \|y-g\| = \inf\limits_{g \in F} \left\|\frac{\bar{x}-f}{\|\bar{x}-f\|}-g\right\| $$

Lets focus on a fixed $g \in F:$

$$= \left\| \frac{\bar{x}}{\|\bar{x}-f\|} -\left( \frac{f}{\|\bar{x}-f\|} + g \right) \right\| =\frac{1}{\|\bar{x} -f\|} \left\| \bar{x}-g' \right\| $$

Where $g'=f+\|\bar{x}-f\|\cdot g \in F$, because its a subspace.

$$\geq \frac{1}{\|\bar{x}-f\|}d(x,F) \geq \frac{1}{\alpha} \geq 1-\varepsilon$$

Where the last two inequalities follow from choice of $f$ and $\alpha$, since g was arbitrary, we conclude:

$$\forall \varepsilon >0 \exists y \in E : \|y\| =1 \text{ and } d(y,F)\geq 1-\varepsilon$$

Answer 2

Since $d(x,F)>0$ for $x\notin F$ and $d(tx,F)=td(x,F)$ there exists $x_1$ with $d(x_1,F)=1$. Choose $a\in F$ with $||x_1-a||$ close to $1$. Now $x=x_1-a$ is almost the point you want...