A functional $T$ on a normed linear space $X$ is said to be Lipschitz provided there is a $c \geq 0$ such that $|T(g) - T(h)| \leq c \|g -h\|$ for all $g, h \in X$ The infimum of such c's is called the Lipschitz constant for $T$. Show that a linear functional is bounded if and only if it is Lipschitz, in which case its Lipschitz constant is $\|T\|_*$
My Try:
$“\Rightarrow"$
Consider $T: X \to \mathbb{R}$ bounded on $X$ $\Leftrightarrow \exists M \geq 0$ such that $|T(f)| \leq M \|f\| $ for all $f \in X$ and the norm $\|T\|_* = M$
Let $f, g \in X$ , by linearity of $T$ $$|T(f) - T(g)| \leq \|T\|_* \|f - g\|$$ Consider $c = \|T\|_*$ $\Rightarrow T$ is Lipschitz
$“\Leftarrow"$
Assume $T$ is not bounded, then there is a $M$ such that $|T(f)| \geq M\|f\|$ and i am stuck
in the second part your assumption that $T$ is not bounded implies that: $$ \forall M \gt 0 \exists f \cdot |Tf| \gt M $$ then you can use Lucas' argument for your contradiction