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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.

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4 Answers 4

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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.

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  • $\begingroup$ Ah, nice! Actually, in the meantime, I found a differentiable example. I'll add it to my question. $\endgroup$ Mar 28, 2011 at 13:54
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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?

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  • $\begingroup$ Sorry Theo; I totally forgot a factor $n^2$ in the definition of $f(x)$ in $B_n$. $\endgroup$
    – Pacciu
    Mar 28, 2011 at 13:54
  • $\begingroup$ No problem. Now the example is perfectly fine. I removed my comment as it is no longer relevant. $\endgroup$
    – t.b.
    Mar 28, 2011 at 13:56
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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.

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Here is another construct. Let $H$ be any Hilbert space and $\{e_n\}_{n=1}^\infty$ be an orthonormal basis. Then $\|e_n - e_m\|^2 = 2$ if $m\not = n$, so $\|e_m - e_n\| = \sqrt{2}$ if $m \not= n$.

Denote by $U_n$ the closed unit ball about $e_n$. Each $U_n$ is closed and is a positive uniform distance ($\sqrt{2} - 1$) from any other $ U_n$; the union of these balls is closed. Put $U = \bigcup_n U_n$. Now define the function $f: U \rightarrow R$ by $f(n) = n$ if $x \in U_n$.

Since $U$ is a closed subset of the ball of radius 2 about the origin in $H$, by the Tietze extension theorem, there is a continuous extension of $f$ defined on the closed ball of radius 2 about the origin. This function is continuous and unbounded.

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  • $\begingroup$ How is that different? You're replacing an easy explicit construction by appealing to a somewhat subtle theorem, but the idea is exactly the same. $\endgroup$
    – t.b.
    Apr 22, 2011 at 3:47

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