[hgibbs]

Depends on what space you define the derivative on. It is of course a bounded (and therefore continuous) operator from C^k to C^{k-1} for any positive integer K.

Additionally, it only really makes sense to talk about bounded operators between topological vectors spaces (as you need to make sense of what it means to be bounded), of which the most commonly dealt with are Banach spaces.

[adament]

It also depends on what topology you put on your space, while the derivative operator is defined on all of C^k, that does not make it continuous. In fact as the GPs example shows, the topology you put on C^k and C^{k-1} must be so that uniform convergence does not imply convergence in C^k (which differs from many peoples intuitive notion of convergence of functions) or cos(x/k) converges to 0 in C^{k-1} (which is just plain weird).

[enriquto]

> Additionally, it only really makes sense to talk about bounded operators between topological vectors spaces

Reading your comment, I wondered how can you define bounded sets in a topological vector space (where you don't have a norm). The definition is cute: a set X if bounded if any neighborhood of 0 can be inflated to include the whole of X.