Prove the following for $T$ a linear functional on $(X,\|·\|)$

(a) If $T$ is a bounded linear functional then $|T(f)|\leq \|T\|_*\|f\|$ for all $f\in X$.

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

(c) A linear function is bounded if and only if it is continuous (Hint: use (b) for the “if" part of the statement)

For a normed linear space $X$, a linear function $T$ on $X$ is said to be bounded if there is an $M≥0$ for which $|T(f)|\leq M\|f\|$ for all $f\in X$.
$\|T\|_*=\inf\{M\geq 0\mid|T(f)|\leq M\|f\|,\text{ any }f\in X\}$

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And the question is? –  Martin Argerami Dec 3 '12 at 8:11
Martin Argerami: The question is to prove (a), (b) and (c) for $T$ a linear functional on $(X,||⋅||)$ –  i_a_n Dec 3 '12 at 8:39
For a) let $M$ be the constant such that $|T(f)|\leq M\|f\|$. Then $\|T\|_\ast \leq M$ by the definition so it remains for you to show $\ge$.
For b): Claim: $\sup\{|T(f)|\mid f\in X,\,\|f\|=1\} = \|T\|_\ast$. Then by a), $|T(f)|\leq \|T\|_*\|f\|$ for all $f$ and hence $\sup\{|T(f)|\mid f\in X,\,\|f\|=1\} \le \|T\|_\ast$. It remains for yout to show $\ge$.
For c): Assume $T$ is bounded, that is, $|T(f)|\leq M\|f\|$ for all $f$ and some $M$. Then $T$ is Lipschitz.
The other direction is the only tricky part of your homework assignment. Assume $T: X \to Y$ is continuous. Fix $\varepsilon>0$ and $x_0$ in $X$. Let $\delta>0$ be such that $T(B_\delta(x_0)) \subset B_\epsilon (T(x_0))$. Pick any $x$ in $X$. Consider $\tilde{x} = \frac{\delta x}{\|x\|} + x_0$. Then $\|Tx\| = \|T(\tilde{x}-x_0)\|\frac{\|x\|}{\delta} \le \varepsilon \frac{\|x\|}{\delta}$.