Dual norm and distance

Question

Let $Z$ be a subspace of a normed linear space $X$ and $x\in X$ has distance $d=\inf\{||z-y||:z\in Z\}$ to $Z$.

I would like to find a function $f\in X^*$ that satifies

$||f||\le1$, $f(x)=d$ and $f(z)=0$

Is it correct that $||f||:=\sup\{|f(x)| :x\in X, ||x||\le 1\}$ because I cannot conclude from this definition that $||f||\le1$

May you could help me with that, thank you very much.

The problem is not well-formed. The definition of dual norm $\|f\|$ that you wrote is correct. However you did not say anything about $f$ other than $f\in X^*$. There is no information to say anything at all about $\|f\|$, $f(x)$, or $f(z)$. What do you know about $f$?
I want to find such a function f that satifies these conditions, sorry if I did not make it clear, so I want to show there exists a function $f\in X^*$ such that $||f||\le 1$, $f(x)=d$ and $f(z)=0$
Please edit your question accordingly. For example, it could say "I would like to find a function $f\in X^*$ such that $\|f\|\le 1$, $f(x)=d$ and $f(z)=0$ for all $z\in Z$."

user53153 · Accepted Answer · 2013-01-10 02:07:11Z

I'll list the ingredients and leave the cooking to you:

The function $d:X\to [0,\infty)$ is sublinear in the sense used in the Hahn-Banach theorem
There is a linear functional $\phi$ on the one-dimensional space $V=\{t x:t\in\mathbb R\}$ such that $\phi(x)=d(x)$ and $|\phi(y)|\le d(y)$ for all $y\in V$. If you get stuck here, mouse over the hidden text.
From 1, 2, and Hahn-Banach you will get $f$.

$\phi(tx)=td(x)$

1 Answer