4
$\begingroup$

I want to find an example of a Banach space $X$ and a continuous map $f:X\rightarrow X$ such that $f$ is not bounded on the unit ball. I do not doubt that such an example exists, but I cannot make it explicit.

$\endgroup$

4 Answers 4

7
$\begingroup$

Find a sequence of disjoint balls $(B_n)$, all contained in the unit ball of $X$ and all having radius greater than some fixed $\epsilon$. Select continuous functions $f_n: B_n \to X$ however you like, such that each $f_n$ is zero outside radius $\epsilon/2$. Then patch together to get a continuous function on $\bigcup B_n$, and extend by zero elsewhere to get a function from $X$ to $X$. There needn't be a uniform bound on the $f_n$ for this to work.

$\endgroup$
2
  • 3
    $\begingroup$ I think it is safer to have the family of balls locally finite too, and not only disjoint. Or also, to have the support of $f_n$ into the ball of radius $\epsilon/2$. Otherwise the resulting glueing function may fail to be continuous. For instance, in $\ell_\infty$ the unit balls centered at $(1+1/n)e_n$ are disjoint and well separated from each other, but any nbd of $0$ meets almost all of them, which allows the glued $f$ to be possibly discontinuous. $\endgroup$ Nov 21, 2014 at 16:31
  • $\begingroup$ Yeah, good point. Corrected. $\endgroup$
    – Nik Weaver
    Nov 21, 2014 at 17:52
5
$\begingroup$

For a somewhat less explicit example, it suffices to find a continuous $g : X \to \mathbb{R}$ which is unbounded on the unit ball (then fix any nonzero $x_0 \in X$ and take $f(x) = g(x) x_0$). Pick a discrete subset $\{x_n\}$ from the unit ball (use the Riesz lemma) and set $g(x_n) = n$. Invoke the Tietze extension theorem to extend $g$ to all of $X$.

$\endgroup$
5
$\begingroup$

For $X=\ell^1$, put $f(x)=\sum_{n\ge 1} nx_i^n$ (this one is even analytic if you understand that word in a not too restrictive sense).

$\endgroup$
2
  • $\begingroup$ Nice. You mean that $f$ is scalar-valued and defined by $\sum_{n\ge 1}nx_n^n$. Your sum on $n$ does not converge, say if $x_1=1$. $\endgroup$
    – Bazin
    Nov 24, 2014 at 20:02
  • $\begingroup$ Yeah, $i$ was a misprint. The general term is $nx_n^n$, just as you wrote. $\endgroup$
    – fedja
    Nov 25, 2014 at 0:55
2
$\begingroup$

On $c_0$ or $\ell^\infty$, let

$$f(x)_j = j \max(0, x_j - 1/2 - \sup_{i\ne j} x_i)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.