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Let $X$ be an infinite-dimensional Banach space, and let $B=\{x\in X: \|x\|\leq 1\}$ be the closed unit ball of $X$. Please give an example of a continuous mapping $F: X\to X$ such that $F(B)$ is unbounded.

Anastasia V
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  • Here is a related question. – David Mitra Jul 08 '15 at 18:34
  • Required to be linear?! – C-star-W-star Jul 09 '15 at 06:30
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    I really don't understand why people vote to close so often for ridiculous reasons: Here as "off-topic" and "unclear what you're asking". I don't have a clean slate either: I voted to close too, but as "duplicate". Someone else gave a link in the comments. I noticed they seem really duplicates, thus voted to close - but only after checking. Moreover if unclear what is asked then be constructive!! Ask in a comment or assume the closest as some guys usually do here on MSE. Folks be more gentle. – C-star-W-star Jul 09 '15 at 06:46

1 Answers1

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EDIT: The following works on any non-reflexive infinite-dimensional Banach space.

Suppose $f$ is a continuous linear functional on $X$ such that $\|f\| = 1$ but $f(x) < 1$ for all $x \in B$, i.e. $f$ does not assume its supremum on $B$. Take $F(x) = x/(1-f(x))$ for $x \in B$, and $F(x) = F(x/\|x\|)$ on its complement.

For example, on the space $c_0$ of sequences converging to $0$ you could take $f(x) = \sum_j f_j x_j$ where $\sum_j |f_j| = 1$ and all $f_j \ne 0$.

EDIT: For a surjective version, let $$F(x) = \dfrac{x}{1-f(x/\max(1,\|x\|)}$$

EDIT: Here's a construction that works in any infinite-dimensional Banach space. There is a sequence $x_n$ such that $\|x_n\| = 1$ and $\|x_i - x_j\| > 1/2$ for $i \ne j$. Thus the closed balls $B_{1/4}(x_n) = \{x: \|x - x_n\| \le 1/4\}$ are disjoint. Define $$ F(x) = \cases{ (1 - 4 \|x - x_n\|) (n-1) x_n + x & if $x \in B_{1/4}(x_n)$\cr x & if $x$ is not in any $B_{1/4}(x_n)$\cr}$$ Note that $F(x_n) = n x_n$ (which makes $F$ unbounded on $B$), while $F(x) = x$ on the boundary of $B_{1/4}(x_n)$.

Robert Israel
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