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A quick question to which I need an answer for my research thesis.

Let $f \colon X \to \mathbb R$ where $X$ is a Banach space and assume that $f \in \mathcal C^0(X)$. This implies that it is locally bounded (it is bounded on bounded sets) as in finite dimension, right?

carlos85
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  • What do you ean by locally bounded ? bounded on compact set ? if yes, then of course it's true... – Surb Sep 14 '22 at 10:40
  • Not compact sets. I mean bounded on bounded sets – carlos85 Sep 14 '22 at 10:42
  • Should follow directly from the definition of the $C^0$ space – daw Sep 14 '22 at 11:08
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    Locally bounded: yes. Bounded on bounded sets: no! See here: https://math.stackexchange.com/questions/1354180/a-continuous-mapping-with-the-unbounded-image-of-the-unit-ball-in-an-infinite-di – PhoemueX Sep 14 '22 at 11:13
  • If $f$ is not linear, then this is likely not true. Here is a counterexample if you allow $f\colon U\rightarrow \mathbb R$ defined only on an open subset $U\subset X$. Let $X$ be the space of continuous functions on the interval $[0,1]$ and let $U$ be the subset of functions that do not have a zero. If $f(x)=x^{-1}$, then $U\cap {\Vert x \Vert_{\infty}\le 1}$ is not mapped to a bounded set. You can probably get something on $U=X$ by modifying this example by suitable cutoff's around the zeros. – Jan Bohr Sep 14 '22 at 11:15

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This is not true if $X$ is infinite-dimensional. Take a countable number of disjoint balls $B(x_n,\epsilon)$ inside the unit ball of $X$. Define any continuous function $f_n$ on $B(x_n,\epsilon)$ that satisfies $f_n(x_n) = n$ and $f|_{\partial B(x_n,\epsilon)} \equiv 0$. Extend each $f_n$ by $0$ to a continuous function on $X$. Now consider $f = \sum\limits_{n = 1}^{\infty} f_n$. Clearly $f$ is continuous as the supports of the summands are disjoint. But $f$ is also unbounded on the unit ball of $X$.

Klaus
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