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) = 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:

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

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:

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

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

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

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

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

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.^[1] Let ${\displaystyle \Omega }$ be an open and connected subset of the complex numbers. Suppose that ${\displaystyle (f_{n})_{n=1}^{\infty ))$ is a sequence of holomorphic functions ${\displaystyle f_{n}:\Omega \to \mathbb {C} }$ that converges uniformly to a function ${\displaystyle f:\Omega \to \mathbb {C} }$ on every compact subset of ${\displaystyle \Omega }$. Then ${\displaystyle f}$ is holomorphic in ${\displaystyle \Omega }$, and moreover, the sequence of derivatives ${\displaystyle (f'_{n})_{n=1}^{\infty ))$ converges uniformly to ${\displaystyle f'}$ on every compact subset of ${\displaystyle \Omega }$.

Theorem.^[2] Let ${\displaystyle \Omega }$ be an open and connected subset of the complex numbers. Suppose that ${\displaystyle (f_{n})_{n=1}^{\infty ))$ is a sequence of univalent^[3] functions ${\displaystyle f_{n}:\Omega \to \mathbb {C} }$ that converges uniformly to a function ${\displaystyle f:\Omega \to \mathbb {C} }$. Then ${\displaystyle f}$ is holomorphic, and moreover, ${\displaystyle f}$ is either univalent or constant in ${\displaystyle \Omega }$.