Local boundedness of continuous functions on a Banach space 
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?
 A: 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.
A: 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.
