Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

Suppose $f \in X^*$, where $X$ is a Banach. Is there such a result: $$\lVert f \rVert_{X^*} = \sup_{x \in X, \lVert x \rVert_X = 1} |l(f,x)|$$ where $l(f, \cdot):X \to \mathbb{R}$ is bounded and linear. here, $l$ is fixed and given. This is true if $l(f,x) = \langle f, x \rangle_{X^*, X}$, for example. Is it true otherwise too?

share|improve this question
What if $l(f,\cdot) = 0$ or $l(f,x) =2 \langle f,x\rangle_{X^\ast,X}$? What kinds of $l$ do you have in mind? That is, what is the motivation for this question? – Martin Feb 12 at 13:02
1  
Your question is not really clear to me but you can define $l(f,x) = f(x)$. And $l(f,\cdot)=f(\cdot)$ is definetely bounded and linear. – tom Feb 12 at 13:03
I guess, any linear operator $l:X^*\otimes X\to\Bbb R$ with $||l||=1$ would be ok.. – Berci Feb 12 at 13:23
1  
@Berci: Certainly not. $l$ could be a rank one operator and still have norm one while $f$ could be in its kernel. If the condition holds for all $f \in X^\ast$ then $l$ must be an isometry $l\colon X^\ast \to X^\ast$, but OP should clarify what $l$ should achieve (is $f$ fixed or not?). _@tom: $\langle f, x\rangle_{X^\ast,X}$ is just another way of writing $f(x)$. – Martin Feb 12 at 13:30
$l$ can really be anything in what I have in mind. Just bounded and linear of the form I wrote above. Yes, $f$ is a fixed functional. – michael_faber Feb 12 at 13:36
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.