Here's an attempt at the second part. I'm much more familiar with the *convenient calculus* of Frölicher, Kriegl, and Michor than with Fréchet derivatives, but for Hilbert spaces, there shouldn't be too much in it.

Let $\rho \colon \mathbb{R} \to \mathbb{R}$ be a smooth bump function at $0$ with support in $[-1/2,1/2]$. Since $\|e_n - e_m\| = \sqrt{2}$ (for $n \ne m$), if we define $f_n \colon H \to \mathbb{R}$ by $f_n(x) = \rho(|e_n - x|^2)$ then the $f_n$s have disjoint support. Then $f := \sum n f_n \colon H \to \mathbb{R}$ is locally smooth and hence smooth. In addition, $f(e_n) = n$ so $f$ maps the unit ball onto an unbounded set.

(How does that look? I'm well aware that I may have overlooked something really obvious!)

*Edit 2012-11-12*: I did overlook something that I shouldn't have done. The argument above is not quite correct: that the $f_n$s have disjoint support is not enough to know that the sum $\sum n f_n$ is smooth. For that, I need to know that the $f_n$s have *locally finite* support. Fortunately, this is true. If $f_n(x) \ne 0$ then $|e_n - x| < 1/4$. Thus for $y \in H$ consider the ball of radius $1/4$ about $y$. If we have $x_1$ and $x_2$ in this ball and $n,m$ such that $f_n(x_1) \ne 0$ and $f_m(x_2) \ne 0$ then $|e_n - e_m| \le |e_n - x_1| + |x_1 - y| + |y - x_2| + |x_2 - e_m| < 1$ whence $n = m$. Hence at most one $f_n$ has support intersecting this ball about $y$ and so the supports of the $f_n$ are locally finite.

closed and boundedis equivalent tocompactness(this property actuallycharacterisesfinite-dimensionality). So in any infinite-dimensional Banach space, closed and bounded is a far weaker property than compactness, so there's no reason why $C^\infty$ functions should have bounded range on a closed, bounded set. I'm sure there are many $C^\infty$ functions with unbounded range on a closed bounded set (it's only thelinearityof linear operators which forces bounded range, nothing to do with $C^\infty$). I'll try to think of some on $\ell^2$. – Zen Harper Aug 17 '10 at 9:42