# Infinite-dimensional normed spaces and the distance

By $X$ we denote an infinite-dimensional normed space (it seems to be obvious that the case of finite dimension is not suitable). Let $X_0$ be a closed subspace of $X$ and $x\in X$. Then there is the distance $d(x,X_0)$ between $x$ and $X_0$ defined as $\inf\{||x-t||:t\in X_0\}$. It is easy to see that $X_0$ is not compact subspace, hence we cannot state that $\exists x_0\in X_0$ $d(x,x_0)=d(x,X_0)$. So, could you help me to build such example?

• $X_0$ is supposed to be closed only. Why should it be compact? – Marc Palm Oct 17 '13 at 14:17
• An example of what? A space where every closed subspace has such an $x_0$? Or a space with a closed subspace $X_0$ with no such best approximation? – Prahlad Vaidyanathan Oct 17 '13 at 14:18
• An example of a space $X$, closed subspace $X_0$ and only one point $x\in X_0$ – user74574 Oct 17 '13 at 14:28
• I mean I just want to get an example when the distance is unreachable – user74574 Oct 17 '13 at 14:35
• What is a subspace? Is is linear or just a subset? – Moishe Cohen Oct 17 '13 at 14:46

Such minimizer exists and is unique if $X$ is uniformly convex and complete. On the other hand, in every non-reflexive Banach space there is a closed subspace for which minimizer does not exist. Lastly, in a reflexive Banach space the minimizer always exists but need not be unique.