Let $E$ be a normed space, $(X,d)$ a metric space and $A \subset E$ a convex bounded set. Show that $f(A)$ is bounded. 
Let $E$ be a normed space, $(X,d)$ a metric space and  $A \subset E$ a convex bounded set. Let $f : A \to X$ be uniformly continuous. Show that $f(A)$ is bounded.

I'm trying to find the right definitions to use here. If $A$ is bounded then for every $x,y \in A$ we have that $[x,y] \in A$. Also  if $A$ is bounded then $d(A) = \sup\{d(x,y) \mid x,y \in A \} < \infty$. So  what I want to show is that $d(f(A)) = \sup\{d(x,y), \mid x,y \in f(A) \} < \infty$? I'm not sure what I should do with the fact that $A$ is convex? From uniform continuity I get that all close points stay close to each other, but not sure that is of help?
 A: Hint: Fix a point $x_0\in A$. Then we have for all $x\in A$ and all $N\in \mathbb{N}_{>0}$
$$ d(f(x_0), f(x)) \leq \sum_{j=1}^N d\left( f(x_0+\frac{j-1}{N}(x-x_0)), f(x_0+\frac{j}{N}(x-x_0)) \right).$$
Now use the fact that $\vert x -x_0 \vert \leq C<\infty$ and the uniform continuity to get an estimate for $N$ sufficiently large.
Here I used the convexity to connect $x_0$ to $x$ by a straight line, which lies completely in $A$.
For general bounded sets this will not work. Take $E=\ell^2(\mathbb{N}, \mathbb{R})$ and $A= \{ e_j : \ j\in \mathbb{N} \}$, where $e_j=(0, \dots, 0, 1, 0, \dots)$ (the $1$ is the $j$th entry). Then any map is uniformly continuous (simply, because $\Vert e_j - e_k\Vert = \sqrt{2}$ for $j\neq k$). Clearly there are unbounded maps $f: A \rightarrow X$ if $X$ is unbounded (pick your favourite one to get a counterexample). Note that this is kind of the easiest example one can get as for finite dimensional normed spaces we have that bounded & closed sets are compact. In that setting we can continuously extend $f$ to the closure of $A$, however, as continous functions map compact sets to compact sets and in metric spaces compact sets are bounded, we get that $f(A) \subseteq f(\overline{A})$ is bounded.
