# Finding a vector in a n.l.s.

Let $X$ be a normed linear space and $Y$ a closed proper subspace. Prove that for all $\varepsilon > 0$, there is an $x \in X$ with $\|x\| = 1$ and such that $\|x − y\| ≥ 1 − \varepsilon$ for all $y \in Y$ .

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## 1 Answer

This is a basic result in functional analysis sometimes called Riesz's Lemma. You should be able to find the statement and proof in most functional analysis texts. For instance this is the Lemma on page two in Joseph Diestel's Sequences and Series in Banach Spaces. A Google search reveals a standard (it seems) proof of this result on page 10 here.

If you just want a (big) hint:

Assume $\epsilon\in(0,1)$ (the result is trivial for $\epsilon\ge1$). Pick any $x$ in $X\setminus Y$. Then the distance $d$ from $x$ to $Y$ is positive (we use the hypothesis that $Y$ is closed here). Thus, you may choose a $z\in Y$ with $0\lt\Vert x-z\Vert <{d\over 1-\epsilon}$. Consider the vector $x-z\over\Vert x-z\Vert$.

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