TITLE: Nonlinear function continuous but not bounded
QUESTION [20 upvotes]: I would like an example of a map $f:H\rightarrow R$, where $H$ is a (infinite dimensional) Hilbert space, and $R$ is the real numbers, such that $f$ is continuous, but $f$ is not bounded on the close unit ball $\{ x\in H : \|x\| \leq 1\}$.
Actually, $H$ could be replaced by any Banach space (but not just a normed space-- that's too easy).  My motivation is that if $f$ is linear, this is impossible; but I have next to no intuition about non-linear functions.
Edit: Here's an example for $c_0$ which is even differentiable (disclaimer: I found it here: http://www.ms.uky.edu/~larry/paper.dir/korea.ps).  Define $f:c_0\rightarrow F$ (where F is your field, real or complex) by $$ f(x) = \sum_{n=1}^\infty x_n^n \qquad (x=(x_n)). $$  You can estimate the sum by a geometric progression, so it does converge.  A bit of checking shows that f is Frechet differentible (so certainly continuous).  But $f(1,1,\cdots,1,0,\cdots)=n$ (if there are $n$ ones) so $f$ is not bounded on the closed unit ball.  What I don't immediately see is how to adapt this to $\ell^2$, say.

REPLY [2 votes]: The example that you gave for $c_0$ can be slightly modified to give a map that works for the $\ell^p$  also.
$$(a_n) \mapsto \sum_n n a_n^n$$
is continuous on $c_0$ with the $\sup $  norm so also on all the $l_p$ with their norm. But $e_n \mapsto n$ so the map is not bounded on the unit ball.
If the Banach space has a Schauder base $e_n$, $(\|e_n \| = 1$), then we have a similar map
$$\sum_n a_n e_n \mapsto \sum_n n a_n^n$$
It would be interesting to provide "direct" examples for other concrete Banach spaces, like  $C([0,1])$, or $L^2(\mathbb{R})$, without using bases.
$\bf{Added:}$ Example for $C([0,1])$. Consider a continuous linear functional of norm $1$ which does not achieve its maximum on the unit ball, for example $l(f) = \int_0^{\frac{1}{2}} f(t) d t - \int_{\frac{1}{2}}^1 f(t) dt$, and take $\phi(f) = \frac{1}{1- l(f)}$ on the closed unit ball.
For $c_0$, take $l(a_n) = \sum_{n\ge 1} \frac{a_n}{2^n}$, of norm $1$, but maximum $1$ not achieved on the closed unit ball.
For $\ell^1$, take $l(a_n) = \sum (1-\frac{1}{n}) a_n$, norm $1$, $\ldots$.
Such examples are possible  only for spaces that are not reflexive.
$\bf{Added:}$
For $L^p([0,1])$ ($p\ge 1$) consider the continuous functional ( non-linear) on the closed unit ball
$$f\mapsto T(f)\colon = \int_0^1 t |f(t)|^p d t$$
with supremum $1$, but which is not achieved. Then $f\mapsto \frac{1}{1-T(f)}$ is continuous and unbounded on the closed unit ball.