I am trying to solve the following question but I am constantly obtaining the same incomplete inequality.
Let $f:A\subseteq\mathbb{R}^{n}\rightarrow\mathbb{R}$ such that $f(x)=\sup\{\langle x,a\rangle:a \in A\}$.
Show that $\left\vert f(x)-f(y) \right\vert \leq \left\Vert A \right\Vert\left\vert x-y \right\vert$. Here $ \left\Vert A \right\Vert=\sup\{\left\Vert a \right\Vert:a \in A\}$
In general, if $f_\alpha$ are Lipschitz with rank $L$ then so is $f=\sup_\alpha f_\alpha$.
$f_\alpha(x) \le f_\alpha(y)+ L \|x-y\| \le f(y) + L \|x-y\|$. Now take the $\sup$ on the left hand side to get $f(x)-f(y) \le L \|x-y\|$. Reversing roles of $x,y$ gives the result.
In the above, we have $f_a(x) = \langle x , a \rangle $ and $|f_a(x)| \le \|a\| \|x\| \le \|A\| \| x\|$, hence we have the desired result.
