Uniform Convergence

Before learning about uniform convergence, let us recall and understand a few related terms and concepts, such as sequence, series, the convergence of a sequence, point convergence, etc. Learning these terms will help you to understand the uniform convergence quickly.

Sequence:

An ordered collection of objects in which repetitions are allowed. It comprises members, called elements or terms like a set.

Series:

The cumulative sum of a given sequence of terms.

Learn in detail about Sequence and series here.

Convergence of a sequence:

A sequence {an} is said to be convergent, if limn→∞ (an) = finite. An example of a convergence sequence is given below.

Let {an} = 1/2n, for all n ∈ N

So, limn→∞ (an) = limn→∞ (1/2n) = 1/2 = 1/∞ = 0 = finite

Sequence of functions:

A set of functions fn(x), n = 1, 2,… defined on a common domain D.

Pointwise Convergence:

A sequence of functions fn: X → ℝ, where X is a subset of ℝ, is said to be converges pointwise on X to the function f: X → ℝ if and only if

limn→∞ fn(x) = f(x) ∀ x ∈ R

Here, fn(x) = f1(x), f2(x), f3(x),….

In the same way, a series of functions

\(\begin{array}{l}\sum_{k=1}^{\infty}f_k(x)\end{array} \)
converges pointwise to S(x) on X if and only if

\(\begin{array}{l}\displaystyle \lim_{n \to \infty}\left [ \sum_{k=1}^{n}f_k(x) \right ]=S(x); \forall x\epsilon X\end{array} \)

What is Uniform Convergence?

Uniform convergence can be defined for both sequences of functions and series of functions, as given below.

Uniform Convergence of Sequence

A sequence of functions fn(x); n = 1, 2, 3,…. Is said to be uniformly convergent to f for a set E of values of x, if for each ε > 0, a positive integer N exists such that |fn(x) – f(x)| < ε for n ≥ N and x ∈ E.

Alternatively, we can define the uniform convergence of a sequence of functions, as follows.

A sequence of functions fn(x); n = 1, 2, 3,…. Is said to be converges uniformly to f if and only if;

\(\begin{array}{l}\displaystyle \lim_{n \to \infty}\left ( \displaystyle \sup_{x\epsilon E}|f_n(x)-f(x)| \right )=0\end{array} \)

That means, supx∈E |fn(x) – f(x)| → 0 as n → ∞.

Uniform Convergence of Series

A series of functions ∑fn(x); n = 1, 2, 3,… is said to be uniformly convergent on E if the sequence {Sn} of partial sums defined by

\(\begin{array}{l}\sum_{k=1}^{n}f_k(x) =S_n(x)\end{array} \)
.

Alternatively, we can define the uniform convergence of a series as follows.

Suppose gn(x) : E → ℝ is a sequence of functions, we can say that the series

\(\begin{array}{l}\displaystyle \sum_{k=1}^{\infty}g_k(x)\end{array} \)
converges uniformly to S(x) on E if and only if the partial sum
\(\begin{array}{l}S_n(x)=\displaystyle \sum_{k=1}^{n}g_k(x)\end{array} \)
converges uniformly to S(x) on E.

Below are simple examples of uniform convergence.

For x ∈ [0, 1), the sequence (1/2)x+n converges uniformly

For x ∈ [0, 1), the sequence xn does not converge uniformly

Read more:

Let’s understand this uniform convergence concept in a better way with the help of the solved problems given below.

Uniform Convergence Solved Examples

Example 1:

Show that the sequence of function {fn}, where fn(x) = 1/(x + n) is uniformly convergent in any interval [0, b], b > 0.

Solution:

Given,

fn(x) = 1/(x + n)

Let’s apply the limit for this function.

f(x) = limn→∞ fn(x) = 0 ∀ x ∈ [0, b]

That means the sequence converges pointwise to 0.

For any ε > 0,

|fn(x) − f(x)| = 1/(x + n) < ε if n > (1/ε) − x, which decreases with x, and 1/ε will be the maximum value.

Let N be an integer such that N ≥ 1/ε

Thus, for ε > 0, there exists N such that |fn(x) − f(x)| < ε, for all n ≥ N

Hence, the sequence of function fn(x) = 1/(x + n) is uniformly convergent in any interval [0, b], b > 0.

Example 2:

Prove that xn is not uniformly convergent.

Solution:

Consider the sequence of functions {xn} defined on [0, 1].

Thus, we quickly identified the pointwise limit of this function.

Indeed, when x ∈ (0, 1), xn → 0 as n → ∞ and, when x = 1, xn → 1 as n → ∞.

Here, we can observe that the pointwise limit of the given sequence is the function ψ(x) = 0, x ∈ [0, 1) and ψ(1) = 1.

So, we say that this sequence is not uniform convergent.

For x ∈ [0, 1), xn = |xn − 0| < ε if and only if n > log ε/ log x such that n0(x) > log ε/ log x.

As x comes close to 1, n0(x) becomes unbounded.

Consequently, there is no way to identify an n0 to make |xn − 0| < ε, n ≥ n0, ∀ x ∈ [0, 1).

Therefore, xn is not uniformly convergent.

Hence proved.

Practice Problems

  1. Prove that the sequence {fn}, where fn(x) = x/(1 + nx2), x is real, converges uniformly on any closed interval I.
  2. Find the uniform convergence of fn(x) = ex/n and gn(x) = xn on [0, 1].
  3. Prove that the sequence {fn}, where fn(x) = xn−1 (1 −x) converges uniformly in the interval [0, 1].

Comments

Leave a Comment

Your Mobile number and Email id will not be published.

*

*