Function Inequality

Question

Let $E$ and $F$ be normed vector spaces and $\mathscr{L}(E,F) = \{f:E \rightarrow F \mid f$ is linear and continuous$\}$ be a normed vector space with the norm $\lVert f \rVert = \sup_{|x|=1} \{|f(x)| \mid x \in E\}$.

Why does the inequality $|f(x)| \leq \lVert f \rVert \cdot |x|$ stand for all $x \in E$? It's trivial when $|x| = 1$ but I cannot prove it to other values of $x$.

Answer 1

For $x \neq 0$, you have $$\vert f(x) \vert = \vert f(\vert x \vert \frac{x}{\vert x \vert}) \vert = \vert x \vert f(\frac{x}{\vert x \vert}) \le \vert x \vert \Vert f \Vert$$