Let $X$ a normed linear spaces, $Y \subset X$ a subspace and $z \in X$ an arbitrary point. How can we show that:

$$\text{dist} (z, Y) = \sup \{\psi(z) \ | \ \|\psi\| = 1, \psi \equiv 0 \ \text{on} \ Y\}$$

Here is a definition of the distance.

  • $\begingroup$ People can probably guess what $\phi$ is supposed to be, but you should probably say it explicitly just to be sure. $\endgroup$ – John Moeller Mar 2 '13 at 1:00
  • $\begingroup$ I was confused when I saw it. The problem does not specify it either. Could you post what it is? $\endgroup$ – user44069 Mar 2 '13 at 1:01
  • $\begingroup$ I can guess that it's a linear function of some sort? But what does it mean to take its norm? $\endgroup$ – John Moeller Mar 2 '13 at 1:04
  • $\begingroup$ I thought the same. Some linear function from $X$ to itself I would guess. $\endgroup$ – user44069 Mar 2 '13 at 1:05

I think you need extra conditions:

  1. $Y$ is a closed subspace of $X$
  2. $\psi\in X^*$

Now let's prove it:

  • The following theorem is one of the principal applications of the Hahn-Banach theorem to normed vector space.

If $Y$ is a closed subspace of $X$ and $z\in X$, there exists $\psi\in X^*$ such that $\psi(x)\not=0$ and $\psi|_Y=0$. $\psi$ can be taken to satisfy $\|\psi\|=1$ and $\psi(x)=$ Dist$(z,Y)$

Therefore, $\operatorname{Dist}(z,Y)\le \sup\,\{\psi (z)\ |\ \|\psi\|=1,\ \psi|_Y=0\}$

  • If $\|\psi\|=1$ and $\psi|_Y=0$, for any $y\in Y$ $$ |\psi(z)|=|\psi(z)-\psi(y)|\le\|\psi\|\cdot \|z-y\|=\|z-y\| \ \Rightarrow \ \psi(z)\le \operatorname{Dist}(z,Y) $$ $$ \Rightarrow \quad \sup\,\{\psi (z)\ |\ \|\psi\|=1,\ \psi|_Y=0\}\le \operatorname{Dist}(z,Y). $$

Based on the above two points, this problem can be solved.

  • $\begingroup$ Do you think you could take a look at this one too? It's also mine. I don't know how to continue from where the first answer left me. math.stackexchange.com/questions/318002/… $\endgroup$ – user44069 Mar 2 '13 at 5:28

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