Discontinuous linear map

In mathematics, linear functionals are an important type of "simple" functions, which are often used as approximations to more general functions, see linear approximation. If the space over which such functionals are defined is a real (or complex) normed space, the useful linear functionals are typically continuous. This article will show that given a normed space which is an infinite-dimensional vector space, there always exist a linear functional which is not continuous defined on that space.

Indeed, let X be a normed space over the field K where K=R or K=C, and f:X→K be a linear functional. Using the linearity property, one can check that f is continuous if and only if f maps the ball

$B=\{x\in X:\|x\|\leq 1\}$

to a bounded set in K, that is, if and only if f is a bounded linear operator. This property is in turn equivalent to requiring that there exists a constant M≥0 such that

$|f(x)|\leq M\|x\|\quad \quad (1)$

for all x in X.

If X is an infinite-dimensional vector space, to show the existence of a linear functional which is not continuous then amounts to constructing f such that equation (1) fails to hold. For that, consider a sequence (e_n)_n (n≥1) of linearly independent elements in X. Define T(e_n)=n e_n for each n=1, 2, ... and extend T linearly to the linear span of the set {e₁, e₂, ... }, which amounts to defining

$T(c_{1}e_{1}+c_{2}e_{2}+\cdots +c_{n}e_{n})=c_{1}e_{1}+2c_{2}e_{2}+\cdots +nc_{n}e_{n}$

for all n≥1 and scalars c₁, c₂,..., c_n. Let T take the value zero outside this linear span. One can check that T is a well-defined linear function, and since it clearly fails to satisfy (1), it is not continuous.

The immediate consequence of this example is that the continuous dual of an infinite-dimensional normed space is a strict subset of the algebraic dual (see dual space). And it emphasizes the fact that an extra dose of caution is needed in doing analysis on infinite-dimensional spaces as compared to finite-dimensional ones.