When does continuous (or uniformly continuous ) function between normed linear spaces carries bounded sets to bounded sets ? I know that if $f:\mathbb R^m \to \mathbb R^n$ is continuous then $f$ carries bounded sets to bounded sets . What if we say $X,Y$ are normed linear spaces and $f:X \to Y$ where $f$ is continuous ? Does $f$ still carries bounded sets to bounded sets ? What if we also assume $f$ is uniformly continuous ?  
 A: This is not true, see this counter example:
Take $X=l^1(\mathbb{R})$ and we will define $f: X \to \mathbb{R}$.
First consider the set $C= \{ u; ||u||_{l^{\infty}} < 1\}\cap B_{l^1}(0,1)$
This is a bounded set in $l^1$ and we will show that its image is unbounded.
Let's do the following decomposition of $C$ into disjoint sets:
$C_0= \{ u; ||u||_{l^{\infty}} < \frac{1}{2}\}\cap B_{l^1}(0,1)$
$C_n= \{ u; 1- \frac{1}{n+1} \leq||u||_{l^{\infty}} < 1- \frac{1}{n+2}\}\cap B_{l^1}(0,1)$
This consists of annuluses of the unit ball in the $l^{\infty}$ norm.
Also one point that will prove to be important is that $C$ is an open set of $l^1$ because the injection $l^1 \to l^{\infty}$ is continuous and $C$ is the preimage of the open ball $B_{l^{\infty}}(0,1)$ with the unit open ball in $l^1$
Let's define $f$ now: 
Define $f(u) =1$ for any $u \in l^1 \setminus C$.
The business to define $f$ inside $C$ is a little more involved, it goes this way:
Set $\eta_n$ to be $\eta_n(x) = \frac{2(x-\left(1-\frac{1}{n+1}\right))}{\frac{1}{n+1} - \frac{1}{n+2}}$ for $x \in [1-\frac{1}{n+1}; 1- \frac{1}{n+1} +\frac{\frac{1}{n+1} - \frac{1}{n+2}}{2}]$, $\eta_n(x)=1 \in [1- \frac{1}{n+1} +\frac{\frac{1}{n+1} - \frac{1}{n+2}}{2}; +\infty)$, $\eta_n(x)=0 \in [0;1- \frac{1}{n+1}]$. So $\eta_n$ grows linearly from $0$ to $1$ between $1-\frac{1}{n+1}$ and the midpoint of $[1-\frac{1}{n+1};1-\frac{1}{n+2}]$ and then it stays constant equal to $1$.
Now define $f$ in $C_n$ by this $f(u)=\frac{1}{1-\left(1-\eta_n(||u||_{l^{\infty}})\right)u_{n-1} - \eta_n(||u||_{l^{\infty}})u_n}$
Let's see first that $f$ is unbouded on $C$: consider $u=(0,0,..,1-\frac{1}{n+1}+\frac{\frac{1}{n+1} - \frac{1}{n+2}}{2} ,0,..)$ where it is nonzero at the $nth$ position. Then $u \in C_n$ and $f(u)>n+1$. That proves the claim.
Now let's prove continuity:
First step: the connection of $f$ between $C$ and $l^1\setminus C$ makes it continuous.
One can see that $\{u , ||u||_{l^{\infty}} = 1 \}$ contains $\partial C$, therefore take a sequence $u^k \to u$ where $u$ is in $\partial C$ (in particular $||u||_{l^\infty} = 1$ ) and $u^k \in C$.
Since $u \in l^1$, $u_n \to 0$ and since $u^k$ converges to $u$ uniformly this will imply that $f(u^k) \to 1$ because $u^k_{n_k} \to 0$ and $u^k_{n_k -1}\to 0$. Where $n_k$ is such that $ u^k \in C_{n_k}$
Last step : $f$ is continuous at the interface between $C_n$ and $C_{n+1}$.
Take $u \in \partial C_n \cap \{ ||u||_{l^{\infty}}= 1-\frac{1}{n+2} \}$, then $u \in C_{n+1}$ and you will have $1-\eta^1_{n+1}(||u||_{l^{\infty}}) = 1$ so that $f(u)$ uses only the coordinate $u_n$ of $u$.
If $u^k \to u$ then there is also uniform convergence, and we have therefore the following:
$||u^k||_{l^{\infty}} \to ||u||_{l^{\infty}}=1-\frac{1}{n+2}$ and $u^k_n \to u_n$.
This implies $1-\eta_n(||u^k||_{l^{\infty}}) \to 0$ and $f(u^k) \to f(u)$
This ends the proof.
