Uniform limit theorem

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

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


More precisely, let X be a topological space, let Y be a metric space, and let ƒn : 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 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 : [0, 1] → R be the sequence of functions ƒn(x) = xn. Then each function ƒn 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 image to the right.

In terms of function spaces, the uniform limit theorem says that the space C(XY) 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(XY) is itself a complete metric space. In particular, if Y is a Banach space, then C(XY) 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 : X → Y is a sequence of uniformly continuous functions converging uniformly to a function ƒ, then ƒ must be uniformly continuous.


In order to prove the continuity of f0, 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_0(x),f_0(y)) < \epsilon, \qquad\forall y \in U

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

d_Y(f_N(t),f_0(t)) < \frac{\epsilon}{3}, \qquad\forall t \in X

Moreover, since fN 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{\epsilon}{3}, \qquad\forall y \in U

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

  d_Y(f_0(x),f_0(y)) & \leq d_Y(f_0(x),f_N(x)) + d_Y(f_N(x),f_N(y)) + d_Y(f_N(y),f_0(y)) \\
                     & < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon, \qquad \forall y \in U

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