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.

A linear map from a finite-dimensional space is always continuous

Let X and Y be two normed spaces and $f:X\to Y$ a linear map from X to Y. If X is finite-dimensional, choose a basis $\left(e_{1},e_{2},\ldots ,e_{n}\right)$ in X which may be taken to be unit vectors. Then,

$$f(x)=\sum _{i=1}^{n}x_{i}f(e_{i}),$$

and so by the triangle inequality,

$$\|f(x)\|=\left\|\sum _{i=1}^{n}x_{i}f(e_{i})\right\|\leq \sum _{i=1}^{n}|x_{i}|\|f(e_{i})\|.$$

Letting

$$M=\sup _{i}\{\|f(e_{i})\|\},$$

and using the fact that

$$\sum _{i=1}^{n}|x_{i}|\leq C\|x\|$$

for some C>0 which follows from the fact that any two norms on a finite-dimensional space are equivalent, one finds

$$\|f(x)\|\leq \left(\sum _{i=1}^{n}|x_{i}|\right)M\leq CM\|x\|.$$

Thus, $f$ is a bounded linear operator and so is continuous. In fact, to see this, simply note that f is linear, and therefore $\|f(x)-f(x')\|=\|f(x-x')\|\leq K\|x-x'\|$ for some universal constant K. Thus for any $\epsilon >0,$ we can choose $\delta \leq \epsilon /K$ so that $f(B(x,\delta ))\subseteq B(f(x),\epsilon )$ ($B(x,\delta )$ and $B(f(x),\epsilon )$ are the normed balls around $x$ and $f(x)$), 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.

A concrete example

Examples of discontinuous linear maps are easy to construct in spaces that are not complete; on any Cauchy sequence $e_{i}$ of linearly independent vectors which does not have a limit, there is a linear operator $T$ such that the quantities $\|T(e_{i})\|/\|e_{i}\|$ grow without bound. In a sense, the linear operators are not continuous because the space has "holes".

For example, consider the space X of real-valued smooth functions on the interval [0, 1] with the uniform norm, that is,

$$\|f\|=\sup _{x\in [0,1]}|f(x)|.$$