Uniform limit theorem

In mathematics, the uniform limit theorem states that the uniform limit of any sequence of continuous functions is continuous.

Counterexample to a strengthening of the uniform limit theorem, in which pointwise convergence, rather than uniform convergence, is assumed. The continuous green functions ${\displaystyle \scriptstyle \scriptstyle \sin ^{n}(x)}$ converge to the non-continuous red function. This can happen only if convergence is not uniform.

Statement

More precisely, let X be a topological space, let Y be a metric space, and let ƒ_n : X → Y be a sequence of functions converging uniformly to a function ƒ : X → Y. According to the uniform limit theorem, if each of the functions ƒ_n is continuous, then the limit ƒ must be continuous as well.

This theorem does not hold if uniform convergence is replaced by pointwise convergence. For example, let ƒ_n : [0, 1] → R be the sequence of functions ƒ_n(x) = xⁿ. Then each function ƒ_n is continuous, but the sequence converges pointwise to the discontinuous function ƒ that is zero on [0, 1) but has ƒ(1) = 1. Another example is shown in the adjacent image.

In terms of function spaces, the uniform limit theorem says that the space C(X, Y) of all continuous functions from a topological space X to a metric space Y is a closed subset of Y^X under the uniform metric. In the case where Y is complete, it follows that C(X, Y) is itself a complete metric space. In particular, if Y is a Banach space, then C(X, Y) is itself a Banach space under the uniform norm.

The uniform limit theorem also holds if continuity is replaced by uniform continuity. That is, if X and Y are metric spaces and ƒ_n : X → Y is a sequence of uniformly continuous functions converging uniformly to a function ƒ, then ƒ must be uniformly continuous.

Proof

In order to prove the continuity of f, we have to show that for every ε > 0, there exists a neighbourhood U of any point x of X such that:

${\displaystyle d_{Y}(f(x),f(y))<\epsilon ,\qquad \forall y\in U}$

Consider an arbitrary ε > 0. Since the sequence of functions {f_n} converges uniformly to f by hypothesis, there exists a natural number N such that:

${\displaystyle d_{Y}(f_{N}(t),f(t))<{\frac {\epsilon }{3}},\qquad \forall t\in X}$

Moreover, since f_N is continuous on X by hypothesis, for every x there exists a neighbourhood U such that:

${\displaystyle d_{Y}(f_{N}(x),f_{N}(y))<{\frac {\epsilon }{3}},\qquad \forall y\in U}$

In the final step, we apply the triangle inequality in the following way:

${\displaystyle {\begin{aligned}d_{Y}(f(x),f(y))&\leq d_{Y}(f(x),f_{N}(x))+d_{Y}(f_{N}(x),f_{N}(y))+d_{Y}(f_{N}(y),f(y))\\&<{\frac {\epsilon }{3}}+{\frac {\epsilon }{3}}+{\frac {\epsilon }{3}}=\epsilon ,\qquad \forall y\in U\end{aligned}}}$

Hence, we have shown that the first inequality in the proof holds, so by definition f is continuous everywhere on X.

References

• James Munkres (1999). Topology (2nd ed.). Prentice Hall. ISBN 0-13-181629-2.