# Why does the norm of a linear functional $T$ satisfy $\|T\|_*=\sup\{|T(f)|\mid f \in X, \|f\|\leq 1\}$?

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? $$\|T\|_*=\sup\{|T(f)|:f \in X, \|f\|\leq 1\}$$

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 $N=sup\{\|T(f)\|, \|f\|=1\}\leq M$
• 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 $\|f\|=1$. – Tsemo Aristide Jun 1 '16 at 2:15
• If $\|f\|<1,$\|T(f)\|$<{1\over{\|f\|}}\|T(f)\|=\|T({f\over{\|f\|}})\leq N$. This implies that $sup\{\|T(f)\|, \|f\|\leq 1\}=sup\{\|T(f)\|, \|f\|=1\}$. – Tsemo Aristide Jun 1 '16 at 2:21