Let be X,Y metric spaces, dX,dY the metrics of X and Y. Let (fn) be a sequence of functions fn: X -> Y.
(fn) is called uniformly converging to a function f: X -> Y iff
for any u > 0, u of R there exists a n > 0, n of N with: for all x of X and all m > n dY( fm(x) , f(x) ) < u

Examples: The sequence of functions (1/n * cos(x) ) converges uniformly to the constant function f(x)=0.
The sequence of functions (xn) doesn't converge uniformly on the closed intervall (0,1)

Introduction and definition

The easiest way of defining convergence of a sequence of functions of fn : S → R defined on some set S is to say that fn → f as n → ∞ iff fn(x) -> f(x) for all x ∈ S. We simply use our definition of convergence of sequences of points for all points in S, and accordingly we say that f is the pointwise limit of the sequence fn.
For reasons that will be explained below, however, we often need the stronger notion of uniform convergence.


A sequence of functions fn : S → R is said to converge uniformly to f : S → R if given any ε > 0 there is a N such that

|fn(x) - f(x)| < ε for all x ∈ S, n > N

If we write out in detail the definition for pointwise convergence we see that the only difference between the two forms of convergence is that for uniform convergence the value of N is not allowed to depend on x, so that the convergence to f is in a sense uniform on S.
Note that uniform convergence therefore implies pointwise convergence. It follows that if we want to see whether a sequence of functions converge uniformly the only candidate for a uniform limit is their pointwise limit.
For testing whether the convergence of a sequence of functions is uniform it is often useful to reformulate the definition.

Rephrased definition:

fn → f uniformly iff

supx∈S |fn(x) - f(x)| → 0 as n → ∞

As usual we treat a series in terms of its partial sums, so a series Σ fn is said to converge uniformly if the sequence of partial sums converges uniformly. A sufficient condition for a series to be uniformly convergent is given by the Weierstrass M-test.


The reason why we want stronger than mere pointwise convergence is that the pointwise limit f of a sequence of functions does not necessarily "inherit" nice properties of the functions fn. Consider the following examples.


These functions fn : [0, 1] → R converge pointwise to f.

a) fn(x) = xn, f(x) = 0 for x ∈ [0, 1), f(1) = 1. This shows that the pointwise limit of continuous functions is not necessarily continuous.
b) Let rk be an enumeration of the rational numbers in [0, 1], fn(x) = 1 if x = rk for some k ≤ n, fn(0) otherwise, and f(x) = 1 if x ∈ Q, f(x) = 0 otherwise. This shows that the limit of Riemann integrable functions is not necessarily Riemann integrable.
c) fn(x) = n for x ∈ (0, 1/n], fn(x) = 0 otherwise, f(x) = 0. This shows that integral of the limit of integrable functions (even if it exists) is not necessarily equal to the limit of the integrals.

This lack of "inheritance" is what limits the usefulness of pointwise convergence. In contrast we have theorems that say that continuity and Riemann integrals are "inherited" by uniform limits.


If fn : S → R are continuous and fn → f uniformly then f is continuous.


For any x ∈ S let ε > 0 be given. By the triangle inequality

|f(x) - f(t)| ≤ |f(x) - fn(x)| + |fn(x) - fn(t)| + |fn(t) - f(t)| for all t ∈ S, n ∈ N

By uniform convergence there is an n such that both the first and last terms of the right hand side of the inequality are < ε/3. By continuity of fn at x there is a neighbourhood U of x such that the middle term is < ε/3 for t ∈ U. Hence

|f(x) - f(t)| < ε for t ∈ U so f is continuous at x.


If fn : [a, b] → R are Riemann integrable and fn → f uniformly then f is Riemann integrable and

∫ fn(x)dx → ∫ f(x)dx as n → ∞

(these and all subsequent integrals are taken from a to b).


Let j, J be the supremum of the lower sums and infimum of the upper sums of f that appear in the definition of the Riemann integral, respectively. Let εn = supx∈[a,b] |fn(x) - f(x)|. Then

fn(x) - εn ≤ f(x) ≤ fn(x) + εn for all x ∈ [a, b], n ∈ N
∫ fn(x)dx + εn(b-a) ≤ j ≤ J ≤ ∫ fn(x)dx + εn(b-a) for all n ∈ N
0 ≤ J - j ≤ 2εn(b-a) for all n ∈ N

As n → ∞ the right hand side → 0, so j = J. Thus f is Riemann integrable, and from the intermediate steps we see that

∫ f(x)dx = J = limn→∞ ∫ fn(x)dx

Of these two theorems it is actually only the first one that is particularly useful. The result on integrability is less interesting since for serious work we do not use the Riemann definition of integrals anyway. Instead we use Lebesgue integrals. For these we have results on integrability under much weaker conditions than uniform convergence, such as the monotone convergence theorem and dominated convergence theorem.


We can naturally generalise the concept of uniform convergence to sequences of functions that map to any metric space rather than just R. The result that the uniform limit of continuous functions is continuous still holds, and the proof is unchanged except for the notation.


A sequence of functions fn : S → X, where (X, d) is a metric space, is said converge uniformly to f if given ε > 0 there is an N such that

d(fn(x), f(x)) < ε for all x ∈ X, n > N


If fn : S → X are continuous and fn → f uniformly then f is continuous.

If we want to view the concept of uniform convergence more abstractly we can note that ||f|| = supx∈X|f(x)| is a norm on the space of continuous functions X → R, provided that X is compact. This norm is called the uniform norm. Uniform convergence is then the same thing as convergence with respect to the uniform norm in this space.

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