# 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 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) = x$^n$. 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:
    $d_{Y}(f(x),f(y))<\varepsilon ,\qquad \forall y\in U$
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:
    $d_{Y}(f_{N}(t),f(t))<{\frac {\varepsilon }{3}},\qquad \forall t\in X$
Moreover, since f$_N$ is continuous on X by hypothesis, for every x there exists a neighbourhood U such that:
    $d_{Y}(f_{N}(x),f_{N}(y))<{\frac {\varepsilon }{3}},\qquad \forall y\in U$
In the final step, we apply the triangle inequality in the following way:
    ${\begin{aligned}d_{Y}(f(x),f(y))&\leq d_{Y}(f(x),f_{N}(x))+d_{Y}(f_{N}(x),f_{N}(y))+d_{Y}(f_{N}(y),f(y))\\&<{\frac {\varepsilon }{3}}+{\frac {\varepsilon }{3}}+{\frac {\varepsilon }{3}}=\varepsilon ,\qquad \forall y\in U\end{aligned}}$
Hence, we have shown that the first inequality in the proof holds, so by definition f is continuous everywhere on X.

## Uniform limit theorem in complex analysis

There are also variants of the uniform limit theorem that are used in complex analysis, albeit with modified assumptions.
Theorem. Let $\Omega$ be an open and connected subset of the complex numbers. Suppose that $(f_{n})_{n=1}^{\infty }$ is a sequence of holomorphic functions $f_{n}:\Omega \to \mathbb {C}$ that converges uniformly to a function $f:\Omega \to \mathbb {C}$ on every compact subset of $\Omega$. Then $f$ is holomorphic in $\Omega$, and moreover, the sequence of derivatives $(f'_{n})_{n=1}^{\infty }$ converges uniformly to $f'$ on every compact subset of $\Omega$.
Theorem. Let $\Omega$ be an open and connected subset of the complex numbers. Suppose that $(f_{n})_{n=1}^{\infty }$ is a sequence of univalent functions $f_{n}:\Omega \to \mathbb {C}$ that converges uniformly to a function $f:\Omega \to \mathbb {C}$. Then $f$ is holomorphic, and moreover, $f$ is either univalent or constant in $\Omega$.