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:

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:

Moreover, since *f _{N}* 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*.

## References

- James Munkres (1999).
*Topology*(2nd ed.). Prentice Hall. ISBN 0-13-181629-2.