From the wiki article on the dual of a norm:
$X$ and $Y$ are normed spaces, and we associate with each $f\in L(X,Y)$ (the space of bounded linear operators from $X$ to $Y$) the number $$\|f\| = \sup\{|f(x)|:x\in X, \|x\| \leq 1\}.$$
At some point in the proof that $L(X,Y)$ is bounded, it seems like they use the inequality $$\|fx\| \leq \|f\|\cdot\|x\|,$$ where $\|x\|\leq 1$.
I can't see why this should holds.
$\|f|\|$ is the supremum of what $|f(x)|$ can be, given that $\|x\|\leq 1$. So, given that $\|x\| \leq 1$, $|fx|$ should be bounded by $\|f\|$. But why should it be bounded by $\|f\|\cdot\|x\|$?
The inequality is clearly true if $x=0$. If $x\not=0$ write $$x=||x||\dfrac{x}{||x||}=||x||\cdot y$$ where $y=x/||x||$ has norm 1. The definition of the operator norm gives that $$||fy||\leq ||f||$$ hence $$\left|\left|f\left(\dfrac{x}{||x||}\right)\right|\right|\leq ||f|| \Rightarrow ||fx||\leq ||f||\cdot||x||$$ by linearity.