Uniform limit theorem


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

Statement

More precisely, let X be a topological space, let Y be a metric space, and let ƒn : XY be a sequence of functions converging uniformly to a function ƒ : XY. According to the uniform limit theorem, if each of the functions ƒn 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 ƒn : → R be the sequence of functions ƒn but has ƒ = 1. Another example is shown in the adjacent image.
In terms of function spaces, the uniform limit theorem says that the space C of all continuous functions from a topological space X to a metric space Y is a closed subset of YX under the uniform metric. In the case where Y is complete, it follows that C is itself a complete metric space. In particular, if Y is a Banach space, then C 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 ƒn : XY 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:
Consider an arbitrary ε > 0. Since the sequence of functions converges uniformly to f by hypothesis, there exists a natural number N such that:
Moreover, since fN is continuous on X by hypothesis, for every x there exists a neighbourhood U such that:
In the final step, we apply the triangle inequality in the following way:
Hence, we have shown that the first inequality in the proof holds, so by definition f is continuous everywhere on X.