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For a normed linear space $X$, a linear functional on $X$ is said to be bounded provided there is an $M \geq 0$ for which $|T(f)|\leq M\|f\|$ for all $f \in X$. Denote $\|T\|_*$ the infinmum of all such $M$.

Why is the following equality true? $$\|T\|_*=\sup\{|T(f)|:f \in X, \|f\|\leq 1\}$$

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Let $N=sup\{\|T(f)\|, \|f\|=1\}$. Let $f\neq 0$, $\|T({f\over\|f\|})\|\leq N$ since $T$ is linear, we deduce $\|T(f)\|\leq N\|f\|$ thus $N\geq M$.

On the other hand,for every $f$ such that $\|f\|=1$, $\|T(f)\|\leq M\|f\|=M$ this implies that $N=sup\{\|T(f)\|, \|f\|=1\}\leq M$

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  • $\begingroup$ Seems like the second half only proves N<=M when ||F||=1 $\endgroup$ – user1559897 Jun 1 '16 at 2:13
  • $\begingroup$ Since you know that $N\geq M$ to show that $N=M$ it is enough to show that $N\leq M$, you can use for that the set of $f$ such that $\|f\|=1$. $\endgroup$ – Tsemo Aristide Jun 1 '16 at 2:15
  • $\begingroup$ i think I see what you meant. Is it right to say that ∥T(f)∥ / ||f|| = ∥T(f/||f||)∥? $\endgroup$ – user1559897 Jun 1 '16 at 2:17
  • $\begingroup$ If $\|f\|<1, $\|T(f)\|$<{1\over{\|f\|}}\|T(f)\|=\|T({f\over{\|f\|}})\leq N$. This implies that $sup\{\|T(f)\|, \|f\|\leq 1\}=sup\{\|T(f)\|, \|f\|=1\}$. $\endgroup$ – Tsemo Aristide Jun 1 '16 at 2:21
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The equality gives you an equivalent definition of the norm of a bounded linear functional. A proof is given here.

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