Does every continuous operator between normed spaces map bounded sets to bounded sets? Suppose you have a continuous operator $A$ between two normed spaces $X$ and $Y$. Does it follow that this operator is bounded in the sense that it takes bounded sets to bounded sets, given that $A$ is not necessarily linear? 
(For linear operators this is true: every bounded set is contained in some ball $\{\|x\|\le R\}$, and then its image is contained in the ball $\{\|x\|\le \|A\|R\}$.)
If not, is linearity the minimum requirement to ensure the boundedness? 
 A: No, it does not. A continuous nonlinear operator can take a bounded set to an unbounded one. I gave an example here, which is also a homeomorphism. Since you don't require $A$ to be a homeomorphism, here's a simpler version: 
Let $\mathcal H$ be a Hilbert space with orthonormal basis $\{e_i : i\in \mathbb N\}$.  Define a nonlinear map $A:\mathcal H\to\mathbb R$ as follows: 
$$A(x)=\sum_{n=1}^\infty n(1-2\|x-e_n\|)^+$$
where $a^+$ means $\max(a,0)$. For each $x$, the sum  has at most  one nonzero term in some neighborhood of $x$, which takes care of the convergence and continuity. 
On the other hand, $A(e_n)=n$, so the image of the closed unit ball is unbounded. 
But no, linearity is not the minimum requirement for sending bounded sets to bounded. For example, Lipschitz continuity is enough. Every continuous linear map is Lipschitz continuous, but  Lipschitz maps need not be linear.  
A: An easy example: 
Consider the continuous function $f(x)=$tan $x$, which maps a bounded set to an unbounded set.
