# Banach space and its closed subspace. a vectors satisfying inequality.

V=a Banach space over R

W=a proper closed subspace of V

Prove : For any $\epsilon > 0$, there is a v $\in$_V_ such that ||v||=1 and ||v+w||$\geq$ $1$ - $\epsilon$

I have shown that there exists such v $\in$ V such that ||v||=$1$

But I am having a trouble to show that ||v||=1 and ||v+w||$\geq$ $1$ - $\epsilon$

I was trying to use the fact that such v with ||v||=$1$ are the vectors on the unit sphere. But I do not know how to proceed with this idea.

So any ideas and opinions would be appreciated.

Thank you.

Let $y \in V \setminus W$ with $\def\norm#1{\left\|#1\right\|}\norm y = 1$. Then, as $W$ is closed $d := \def\dist{\mathop{\rm dist}}\dist(y, W) > 0$. Choose $\delta > 0$ such that $\frac{d}{d+ \delta} > 1-\epsilon$. There is some $w \in W$ with $\norm{y-w} < d + \delta$. We let $v := \frac{y-w}{\norm{y-w}}$. Then $\norm v = 1$ and $$\dist(v,W) = \dist\left(\frac{y-w}{\norm{y-w}}, W\right) = \frac 1{\norm{y-w}} \dist(y-w, W) = \frac 1{\norm{y-w}}\dist(y,W) > \frac{\dist(y,W)}{d+\delta} = \frac{d}{d+\delta} > 1-\epsilon.$$
• Because $W$ is a subspace and for $w' \in W$ $$\| (y-w)/\|y-w\| - w' \| = \frac 1{\|y-w\|} \cdot \| (y-w) - \|y-w\|w' \|.$$ – martini Apr 7 '15 at 8:24