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Question:

Let $x\in X$, $X$ is a normed linear space and let $X^{*}$ denote the dual space of $X$. Prove that$\|x\|=\sup_{\|f\|=1}|f(x)|$ where $f\in X^{*}$.


My proof:

Let $0\ne x\in X$, using HBT take $f\in X^{*}$ such that $\|f\|=1$ and $f(x)=\|x\|$.

Now, $\|x\|=f(x)\le|f(x)|\le\sup_{\|x\|=1}|f(x)|=\sup_{\|f\|=1}|f(x)|$, this implies $\|x\|\le\sup_{\|f\|=1}|f(x)|\quad (1)$

Since $f$ is a bounded linear functional $|f(x)|\le\|f\|\|x\|$ for all $x\in X$. Since$\|f\|=1$, $|f(x)|\le\|x\|$ for all $x\in X$. This implies $\|x\|\ge\sup_{\|f\|=1}|f(x)|\quad(2)$

Therefore $(1)$ and $(2)$ gives $\|x\|=\sup_{\|f\|=1}|f(x)|$.


If $x=0$, the result seems to be trivial, but I am still trying to convince myself. Still I have doubts about my proof, is it correct? Please help.

Edit:

Please note that, I use the result of the one of the consequences of Hahn-Banach theorem. That is, given a normed linear space $X$ and $x_{0}\in X$ $x_{0}\ne 0$, there exist $f\in X^{*}$ such that $f(x_{0})=\|f\|\|x_{0}\|$

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    The $g$ **varies** amongst every element of $X^*$ such that $\|g\| = 1$. While $f$ is a certain carefully chosen. Notice that $f \in \{g \in X^*| \|g\| = 1\}$. Try to figure what you mean when you say that $f(x) = \|x\|$, and what you mean when you say $\sum_{\|f\|=1} |f(x)|$.2011-11-01

2 Answers 2

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To put the discussion in comments to an end:

Yes, your proof is correct (the minor details that were missing—or slightly confusing—were essentially corrected).

There are two ingredients to the proof:

  1. By definition of the operator norm we have $\sup_{f \in X^\ast, \|f\|\leq 1}|f(x)| \leq \|x\|$.

  2. Let $U = \langle x \rangle$ be the subspace generated by $x$. Define a linear functional $\tilde{g}$ on $U$ by setting $\tilde g(tx) = t$ for each scalar $t$. This functional satisfies $\|\tilde{g}\| = 1$ unless we're in the case $x = 0$ in which $\tilde{g} = 0$. By Hahn-Banach we may extend that functional to a linear functional $g$ on all of $X$ such that $\|\tilde{g}\| = \|g\|$. Thus, $\sup_{f \in X^\ast, \|f\|\leq 1}|f(x)| \geq |g(x)| = \|x\|$.

Piecing 1. and 2. together we have $\|x\|\leq \sup_{f \in X^\ast, \|f\|\leq 1}|f(x)| \leq \|x\|$, so we must have equality.

Note that this argument shows in particular that the canonical inclusion $i: X \to X^{\ast\ast}$ given by $x \mapsto i_x$, where $i_x(f) = f(x)$ is an isometric embedding.

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    @t.d: Yes it is very clear.2011-10-31
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Thanks for the comments. Let see....


Let $0\ne x\in X$, using the consequence of HBT (analytic form) take $g\in X^{*}$ such that $\|g\|=1$ and $ g(x)=\|x\|$.

Now, $\|x\|=g(x)\le|g(x)|\le\sup_{\|f\|=1}|f(x)|$, this implies $\|x\|\le\sup_{\|f\|=1}|f(x)|\quad (1)$

Since $f$ is a bounded linear functional (given): $|f(x)|\le\|f\|\|x\|$ for all $x\in X$.

For a linear functional $f$ with $\|f\|=1$ we have by defintion, $|f(x)|\le\|x\|$ for all $x\in X$. This implies $\|x\|\ge\sup_{\|f\|=1}|f(x)|\quad(2)$

Therefore $(1)$ and $(2)$ gives $\|x\|=\sup_{\|f\|=1}|f(x)|$.

If $x=0$, the result is trivial.


Any more comments?

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    @t.b. thanks for your vote, you are a great analyst!2011-11-03