Continuous map but not bounded on Banach space Let $X$ be a Banach space. I am looking for an example of a function $f\in C(X)$ but $f\notin B(X)$, i.e., there exist a bounded set $S\subset X$ such that $f(S)$ is not bounded in $X$. 
 Note that the domain of $f$ is all of $X$.
 A: Consider a sequence of vectors $\{v_i\}_{i\in\Bbb N}$ such that $\lVert v_i\rVert=1$ and $\lVert v_i-v_j\rVert\ge \frac12$ for all $i\ne j$. Such a sequence exists by a corollary of Riesz's lemma. Now, consider the usual bump on $[-1,1]$ \begin{align}\varphi:\Bbb R&\to\Bbb R \\ \varphi(x)&=\begin{cases}e\cdot\exp\frac1{x^2-1}&\text{if }-1<x<1\\ 0&\text{if }x\le-1\lor x\ge1\end{cases}\end{align}
And the function \begin{align}\Phi: X&\to \Bbb R\\ \Phi(x)&=\begin{cases}j\cdot\varphi(5\lVert x-v_j\rVert)&\text{if }\lVert x-v_j\rVert<\frac14\\ 0&\text{if }\forall j,\ \lVert x-v_j\rVert\ge \frac14\end{cases}\end{align}
You can check that it is continuous and that $\Phi[S]$ is unbounded when $S$ is the unit ball (or just $S=\{v_j\,:\, j\in\Bbb N\}$, really).
Added: Ah, you wanted the image to be in $X$; consider $\Phi(x)v$ for some non-zero vector, then.
A: An example with target space $\mathbb R$, but you can easily adapt it.
Sketch: pick $X$ a Hilbert space and $(e_n) $ an infinite orthonormal set. Since $\| e_n-e_m\| =\sqrt{2}$ we can construct continuous bump functions $f_n$ that satisfy: $f_n$ has support in $B(e_n, \sqrt{2} /2)$ and $f_n(e_n) =n$. Now take $f=\sum_{n\geq 1} f_n$. 
