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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?

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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 '13 at 13:02
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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 '13 at 13:03
    
I guess, any linear operator $l:X^*\otimes X\to\Bbb R$ with $||l||=1$ would be ok.. –  Berci Feb 12 '13 at 13:23
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@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 '13 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 '13 at 13:36
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