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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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It will become correst as soon as you justify the existence of the $f$ you use at the very beginning of the proof. What general theorem (or corollary thereof) are you invoking? Also you would want to phrase this first part differently. For instance you could say "let $f\in X^*$ be such that $||f||=1$ and $f(x)=||x||$ : such an $f$ exists by $\dots$ . Then we have $\dots$" and finish your proof. –  Olivier Bégassat Oct 30 '11 at 17:58
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If $f\in X^*$ with $\lVert f\rVert =1$, how do you know that $f(x)=\lVert x\rVert$? (or you have to prove that such a $f$ exists, using Hahn-Banach theorem, but you should mention it) –  Davide Giraudo Oct 30 '11 at 17:58
    
@DavideGiraudo: Forgive me, but I think I do not have to mention the consequence of Hahn-Banach theorem. Of course I use the result of the theorem. –  Hassan Muhammad Oct 30 '11 at 18:03
    
Apart from what has been said : (1) Be careful with variables : you can't define $f$ then write something like $\sup_f |f(x)|$. (2) Something is wrong in this line : $\sup_x |f(x)| = \sup_f |f(x)|$. (3) After equality (1), $f$ has nothing to do with the $f$ from the begining (cf remark (1)). –  user10676 Oct 30 '11 at 18:12
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It si not correct because you say: "let $f \in X^*$ ∗then* $\|f\| = 1$ and $f(x) = \|x\|$". The "conclusion" (what is after the "then") is not correct! The theorem others are pointing you to warrants the existence of such an $f \in X^*$. That is, you should say "using HB theorem, take an $f \in X^*$ such that..." –  André Caldas Oct 30 '11 at 18:31
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2 Answers 2

up vote 5 down vote accepted

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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Note that the comments you received (after you clarified what $f$ is at the beginning of your argument) were essentially notational quibbles. –  t.b. Oct 30 '11 at 21:16
    
From where do you get $g$? –  Hassan Muhammad Oct 31 '11 at 4:05
    
@Hassan: Sorry, I don't understand your question. In case there was a confusion in the notation, I hope it's clearer now. –  t.b. Oct 31 '11 at 6:36
    
@t.d: Yes it is very clear. –  Hassan Muhammad Oct 31 '11 at 16:50
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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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2  
In the second part: instead of "Since $\|f\| = 1$ (given)" I would write: For a linear functional $f$ of norm $\|f\| =1$ we have $|f(x)| \leq \|x\|$ by definition of the operator norm. This implies $(2)$. –  t.b. Oct 31 '11 at 6:40
    
@t.b: What you said is much better. –  Hassan Muhammad Nov 1 '11 at 15:59
    
I think we essentially agree what the "right" proof of this result is. It's ultimately a matter of style but I always prefer to make full and grammatical sentences and indicate how to prove things instead of alluding to results somewhat vaguely, as you do by writing: "using the consequence of HBT (analytic form)" or "Since $f$ is a bounded linear functional (given)". As I wrote in my answer, your proof is perfectly fine from the mathematical point of view. –  t.b. Nov 1 '11 at 19:42
    
@t.b.Thanks for your comment. –  Hassan Muhammad Nov 3 '11 at 10:16
    
you're welcome! I see that no-one's voted for your answer, yet, so here we go... –  t.b. Nov 3 '11 at 10:18
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