= Solution
If $f$ is a <bounded linear operator>, linearity gives $\|f(u)-f(v)\|\leq C\|u-v\|$. It is therefore Lipschitz and continuous everywhere.
Conversely, suppose $f$ is continuous at zero. There is $\delta>0$ such that $\|u\|<\delta$ implies $\|f(u)\|<1$. For $u\ne0$, apply this to $v=\delta u/(2\|u\|)$ and use homogeneity:
$$
\frac{\delta}{2\|u\|}\|f(u)\|=\|f(v)\|<1.
$$
The value at $u=0$ is zero by linearity. Consequently
$$
\boxed{\|f(u)\|\leq\frac2\delta\|u\|\text{ for every }u.}
$$
Thus \b[boundedness and continuity are equivalent for linear maps between normed spaces]. No finite dimensionality or completeness is needed for this equivalence.
Back to article page
