Let$\left(U,\|\cdot\|_{U}\right)$ and $\left(V,\|\cdot\|_{V}\right)$ be normed Vectorspaces and $X: U \longrightarrow V$ linear and continous.
$$ \|X\|:=\sup \left\{\|X(u)\|_{V}: u \in U \text { and }\|u\|_{U}=1\right\} $$
I first proved that $\|X\|$ is finite.
Since $X$ is continuous, it is continuous at $0 \in U$. Therefore there is $\delta>0$ such that $\|u-0\| \leq \delta$ $\implies$ $\|X u\|=\|X u-X 0\| \leq 1 .$ Thus, $\|u\| \leq 1$ implies $\|X u\| \leq 1 / \delta$ and hence $\|X\| \leq 1 / \delta .$ Now I want to show, that $X$ is Lipschitz continuous and $\|X\|$ is the smallest Lipschitz constant. But I am having difficulties applying the Definition: $$ \underset{L \geq 0}{\exists} \underset{x, y \in M}{\forall} \quad d_{Y}(f(x), f(y)) \leq L d_{X}(x, y) $$
I came to the conclusion that I have to proof the following statement, but I am not able to do it. $$ \forall u, w \in U \space|X(u)-X(w)|_{V} \leq L\|u-w\|_{U} \space \implies \|X\| \leq L ). $$
Take $u,w \in U$. If $u \neq w$, then $u-w \neq 0$ and $\left\|\frac{u-w}{\|u-w\|_U} \right\|_U = 1$. Hence $\left\|X\left(\frac{u-w}{\|u-w\|_U}\right) \right\|_V \leqslant \|X\|$, and by linearity, $\|X(u-w)\|_V \leqslant \|X\| \|u-w\|_U$. This last inequality is also true if $u=w$. Hence, $\|X\|$ is a Lipschitz constant.
Now suppose $L$ is a Lipschitz constant. Then for all $u$ with $\|u\|_U = 1$, we have $\|X(u)\|_V= \|X(u-0)\|_V \leqslant L \|u-0\|_U= L$, and by the very definition of $\|X\|$, $L$ is bigger than $\|X\|$.
It follows that $\|X\|$ is a Lipschitz constant and that all Lipschitz constants are bigger than $\|X\|$. The result follows.