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.
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4 Answers
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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.
answered Nov 21, 2014 at 15:57
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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. $\endgroup$ Commented Nov 21, 2014 at 16:31
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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 $X$.
answered Nov 21, 2014 at 15:55
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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).
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$\begingroup$ 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 $x_1=1$. $\endgroup$– BazinCommented Nov 24, 2014 at 20:02
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$\begingroup$ Yeah, $i$ was a misprint. The general term is $nx_n^n$, just as you wrote. $\endgroup$– fedjaCommented Nov 25, 2014 at 0:55
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On $c_0$ or $\ell^\infty$, let
$$f(x)_j = j \max(0, x_j - 1/2 - \sup_{i\ne j} x_i)$$
answered Nov 21, 2014 at 16:26