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Let $X$ be an infinite-dimensional Banach space and $f : X \to \mathbb{R}$ continuous (not necessarily linear).

Can $f$ be unbounded on the unit ball?

Of course, in a locally compact space these are impossible. Since $X$ is not locally compact one would guess these are possible, but I cannot think of an example.

Are such examples possible even when $X$ is, say, separable Hilbert space?

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I removed question 2 because it was silly. Of course a continuous function cannot be unbounded on every ball centered at the origin. On sufficiently small balls it must be bounded between $f(0)-1$ and $f(0)+1$. –  Nate Eldredge Jul 14 '12 at 6:00
    
The Hilbert space case is covered here with essentially the same idea as in Brian's answer. The main point is that you can find countably many disjoint closed balls in the unit ball and then you can scale bump functions in these balls in order to get an unbounded continuous function. –  t.b. Jul 14 '12 at 7:21
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2 Answers

up vote 3 down vote accepted

For the first question, let $X=\ell_\infty$. For $n\in\Bbb N$ let $x_n\in X$ be defined by $$x_n(k)=\begin{cases}1,&\text{if }k=n\\0,&\text{otherwise}\;,\end{cases}$$ and let $$V_n=\left\{y\in X:\|y-x_n\|<\frac14\right\}\;.$$ Let $y\in X$ and $B=\{z\in X:\|y-z\|<1/4\}$, and suppose that $B\cap V_n\ne\varnothing$. Then there is a $z\in X$ such that $\|z-y\|<1/4$ and $\|z-x_n\|<1/4$, so $\|y-x_n\|<1/2$. If $\|y-x_m\|<1/2$ as well, then $\|x_n-x_m\|<1$, and therefore $m=n$. Thus, $\mathscr{V}=\{V_n:n\in\Bbb N\}$ is a locally finite family of open sets.

Let $f_n:X\to\Bbb R$ be any continuous function such that $f(x_n)=n$ and $f(y)=0$ for $y\in X\setminus V_n$. Finally, let $f=\sum_{n\in\Bbb N}f_n$. Since $\mathscr{V}$ is locally finite, $f$ is a continuous function from $X$ to $\Bbb R$, and clearly $f$ is unbounded on the unit ball.

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Thanks, nice and simple. And in fact, we can use Riesz's lemma to construct an analogous sequence $\{x_n\}$ in any infinite-dimensional Banach space, whereupon the rest of the argument follows. (For Hilbert space we could just take the $x_n$ orthonormal.) –  Nate Eldredge Jul 14 '12 at 6:07
    
Very nice example. –  copper.hat Jul 14 '12 at 6:12
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A pseudocompact metric space is compact, so there exist unbounded continuous functions on the closed unit ball of every infinite dimensional Banach space, which can be continuously extended by the Tietze extension theorem.


Here is a suggestion based on a post by Henno Brandsma on Ask a Topologist from Nov 28, 2005. Let $(x_n)$ be a sequence (of distinct points) in the unit ball with no convergent subsequence, let $f(x_n)=n$, which is continuous on the closed and discrete set $\{x_n\}$, and extend by Tietze.

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