Discontinuous linear map

In mathematics, linear maps form an important class of "simple" functions which preserve the algebraic structure of linear spaces and are often used as approximations to more general functions (see linear approximation). If the spaces involved are also topological spaces (that is, topological vector spaces), then it makes sense to ask whether all linear maps are continuous. It turns out that for maps defined on infinite-dimensional topological vector spaces (e.g., infinite-dimensional normed spaces), the answer is generally no: there exist discontinuous linear maps. If the domain of definition is complete, it is trickier; such maps can be proven to exist, but the proof relies on the axiom of choice and does not provide an explicit example. Let X and Y be two normed spaces and a linear map from X to Y. If X is finite-dimensional, choose a basis in X which may be taken to be unit vectors. Then, and so by the triangle inequality, Letting and using the fact that for some C>0 which follows from the fact that any two norms on a finite-dimensional space are equivalent, one finds Thus, is a bounded linear operator and so is continuous. In fact, to see this, simply note that f is linear, and therefore for some universal constant K. Thus for any we can choose so that ( and are the normed balls around and ), which gives continuity. If X is infinite-dimensional, this proof will fail as there is no guarantee that the supremum M exists. If Y is the zero space {0}, the only map between X and Y is the zero map which is trivially continuous. In all other cases, when X is infinite-dimensional and Y is not the zero space, one can find a discontinuous map from X to Y. Examples of discontinuous linear maps are easy to construct in spaces that are not complete; on any Cauchy sequence of linearly independent vectors which does not have a limit, there is a linear operator such that the quantities grow without bound. In a sense, the linear operators are not continuous because the space has "holes".