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) = 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:

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{align}
  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{align}

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 \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.^[2] Let \Omega be an open and connected subset of the complex numbers. Suppose that (f_n)_{n=1}^{\infty} is a sequence of univalent^[3] 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.

Notes

^ Theorems 5.2 and 5.3, pp.53-54 in E. M. Stein and R.Shakarchi's Complex Analysis.
^ Section 6.44, pp.200-201 in E. C. Titchmarsh's The Theory of Functions. Titchmarsh uses the terms 'simple' and 'schlicht' (function) in place of 'univalent'.
^ Univalent means holomorphic and injective.

References

James Munkres (1999). Topology. 2nd ed. Prentice Hall. ISBN 0-13-181629-2.
E. M. Stein, R. Shakarchi (2003). Complex Analysis (Princeton Lectures in Analysis, No. 2), Princeton University Press.
E. C. Titchmarsh (1939). The Theory of Functions, 2002 Reprint, Oxford Science Publications.