Nonlinear function continuous but not bounded

Solution 1:

An easy way of doing it in any infinite-dimensional Banach space is to observe that there is a countable discrete subset $\{x_{n}\}_{n = 1}^{\infty}$ of the unit ball by Riesz's lemma (in the Hilbert case you can simply take any orthonormal system). This means that you can find a sequence or radii $r_{n}$ such that the closed balls $\bar{B}_{2r_{n}}(x_{n})$ are pairwise disjoint (take e.g. $r_{n} = \frac{1}{4}$ in the Hilbert setting - I'm being generous here). Putting $f_{n}(x) = \max{\{0,r_{n} - \|x- x_n\|\}}$ you get a continuous function $f_n$ supported on $\bar{B}_{r_n}(x_n)$. The function $f(x) = \sum_{n=1}^\infty \frac{n}{r_n} f_n(x)$ is unbounded on the unit ball because $f(x_{n}) = n$ and it is clearly continuous since the balls $\bar{B}_{2r_n}(x_n)$ are pairwise disjoint.

I don't know of a "natural" example off the top of my head.

Solution 2:

Can't one reason in the following manner?

Let $H=\ell^2$ and let $\{ e^n:=(\delta_m^n)\}_{n\in \mathbb{Z}}$ be its standard base ($\delta_m^n$ is Kronecker).

Then consider the open balls $B_n:=B(e^n;\frac{1}{2})$: these balls are pairwise disjoint (because the sum of their radii is less than the distance between their centers, which equals $\sqrt{2}$).

In each ball set $f(x):=(n^2+1)(1-2|x-e^n|)$, so that $f(x)$ is radially-decreasing, continuous and bounded in $B_n$ (for the image of $B_n$ is the interval $]0,n^2+1]$) and $f(x)$ approaches zero when $x$ approaches the boundary $\partial B_n$.

Now we have a function $f(x)$ defined on $\bigcup_n B_n$ and we want to extend it to the whole space: the easier way to do this is by setting $f(x)=0$ when $x\notin \bigcup_n B_n$.

The extended function $f(x)$ is defined, continuous and unbounded in $\ell^2$ (for it is unbounded on $B(o;1)$, because $\displaystyle \lim_{n\to \infty}|f(e^n)|= +\infty$), and it is obviously nonlinear: in fact if $R>0$ is sufficiently large, then $f(Re^0+Re^1)=0\neq 3R=Rf(e^0)+Rf(e^1)$ .

It makes sense for me. What do you think?