Advanced Analysis

Uniform Convergence

Video. Uniform Convergence

Some Topics Covered Uniform convergence defined; example and nonexample; the sup norm; continuity of the limit; Cauchy criterion

1. Uniform Convergence Defined

Definition
Let \(E \subset {\mathbb R}\) be a set and let \(\{f_n\}_{n=1}^\infty\) be a sequence of real-valued functions on \(E\) and suppose that \(f\) is also a real-valued function on \(E\).
  • We say that \(\{f_n\}_{n=1}^\infty\) converges pointwise to the function \(f\) on the set \(E\) as \(n \rightarrow \infty\) when for each \(x \in E\) and each \(\epsilon > 0\), there is some \(N \in \mathbb N\) such that \(|f_n(x) - f(x)| < \epsilon\) for all \(n > N\). Here \(N\) is explicitly allowed to depend on \(\epsilon\) and on \(x\).
  • We say that \(\{f_n\}_{n=1}^\infty\) converges uniformly to the function \(f\) on the set \(E\) when for each \(\epsilon > 0\), there is some \(N \in \mathbb{N}\) such that every \(n > N\) and \(x \in E\) satisfy \(|f_n(x) - f(x)| < \epsilon\). Here \(N\) is explicitly allowed to depend on \(\epsilon\) and explicitly does not depend on \(x\). In other words, uniform convergence occurs when the same \(N\) can be used at all points \(x \in E\).
Important
Note that uniform convergence is a strictly stronger notion than pointwise convergence. In particular, uniform convergence always implies pointwise convergence but the converse is not necessarily true.
Example
The functions \(f_n(x) := x^n\) on \([0,1]\) converge pointwise as \(n \rightarrow \infty\) to
\[ f(x) := \begin{cases} 0 & x \in [0,1), \\ 1 & x = 1, \end{cases}\]
but do not converge uniformly.
Proof
Meta (Main Idea)
Certainly \(x^n \rightarrow 0\) for any fixed \(x \in [0,1)\) and \(1^n = 1\) is a constant and therefore convergent sequence. But when \(\epsilon = \frac{1}{2}\), for example, for every \(n\) there is always some point \(x \in [0,1)\) such that \(|x^n - 0| > \frac{1}{2}\). This is an immediate consequence of the fact that \(x^n\) is continuous and equals \(1\) at \(x=1\).

For an explicit example, we can show that \(x = 1 - \frac{1}{2n}\) is a specific point for which \(x^n > \frac{1}{2}\). This is a consequence of the inequality \((1-t)^n \geq 1-nt\) for \(t \in [0,1]\). This fact is easy to establish by induction because it's trivially true when \(n = 0,1\), and in general,
\[{}(1-t)^{n+1}{}\]
\[{}= (1-t)^n (1-t){}\]
\[{}\geq (1-nt) (1-t){}\]
\[{}= 1 - (n+1) t + n t^2{}\]
\[{}\geq 1 - (n+1)t.{}\]
Example
The functions \(f_n(x) = x^n (1-x)\) converge both pointwise and uniformly to the zero function as \(n \rightarrow \infty\).
Proof
Meta (Main Idea)
Pointwise convergence is again a direct application of our basic knowledge regarding sequences of real numbers.

Now
\[ f_n'(x) = n x^{n-1} - (n+1) x^{n}\]
and so \(f'_n(x) = x^{n-1}(n - (n+1) x)\). This means that the only critical point of \(f_n\) on the interval \((0,1)\) is at \(x = \frac{n}{n+1}\). Since \(f_n\) is nonnegative and equals zero at \(x = 0,1\), it follows that the maximum of \(f_n\) on \([0,1]\) must be attained at \(x = \frac{n}{n+1}\) and consequently
\[ 0 \leq f_n(x) \leq \left( \frac{n}{n+1} \right)^n \left( 1 - \frac{n}{n+1} \right). \]
In particular, \(0 \leq f_n(x) \leq \frac{1}{n+1}\) for all \(x \in [0,1]\) and all \(n\). Thus for any \(\epsilon > 0\), if \(N\) is so large that \(\frac{1}{N+1} < \epsilon\), then any \(n > N\) also has \(\frac{1}{n+1} < \epsilon\) and therefore \(0 \leq f_n(x) < \epsilon\) for all \(x \in [0,1]\).

2. Continuity

Theorem
Suppose that \(\{f_n\}_{n=1}^\infty\) is a sequence of real-valued functions on some domain \(E\) which are continuous and converge uniformly to \(f\) as \(n \rightarrow \infty\). Then \(f\) must also be continuous.
Proof
This is an iconic argument known as the “epsilon thirds” argument. We fix \(y \in E\) and estimate \(|f(x) - f(y)|\) by splitting into three distinct parts. Specifically,
\[{}f(x) - f(y){}\]
\[{}= (f(x) - f_n(x)){}\]
\[{}+ (f_n(x) - f_n(y)){}\]
\[{}+ (f_n(y) - f(y)){}\]
for any value of \(n\) that we choose.

Begin by fixing any \(\epsilon > 0\). Uniform convergence implies the existence of \(N\) such that \(|f_n(z) - f(z)| < \epsilon/3\) for all \(z \in E\) whenever \(n > N\). Choosing the specific value \(n := N+1\) and applying the inequality for both \(z = x\) and \(z = y\), we see that for our specific choice of \(n\),
\[ |f(x) - f_n(x)| < \frac{\epsilon}{3}\]
and
\[ |f_n(y) - f(y)| < \frac{\epsilon}{3}\]
for any two points \(x, y \in E\). But for this specific \(n\) and any fixed \(y \in E\), \(f_n\) is continuous at \(y\), so for any \(\epsilon > 0\), there is some \(\delta > 0\) such that every \(x \in E\) with \(|x-y| < \delta\) has \(|f_n(x) - f_n(y)| < \epsilon/3\). It follows that every \(x \in E\) with \(|x-y| < \delta\) must have
\[{}|f(x) - f(y)|{}\]
\[{}= |f(x) - f_n(x)|{}\]
\[{}+ |f_n(x) - f_n(y)|{}\]
\[{}+ |f_n(y) - f(y)|{}\]
\[{}< \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3}{}\]
\[{}= \epsilon.{}\]
Corollary
If a sequence of continuous functions \(\{f_n\}_{n=1}^\infty\) converges pointwise to a discontinuous function as \(n \rightarrow \infty\), the convergence cannot be uniform.

This corollary applies directly to the example \(f_n(x) := x^n\) on \([0,1]\) already analyzed above.

3. The Sup Norm

Definition
Given an interval \(I\) and a real function \(f\) on \(I\), we define the
uniform or supremum norm of \(f\) to equal the quantity
\[ ||f||_{U(I)} := \sup_{t \in I} |f(t)|. \]
(Note that there are many other notations you may encounter for this same concept.)

3.1. Basic Properties of the Sup Norm

  • By the Extreme Value Theorem, the supremum of any continuous function on a compact interval is always attained, so if \(I\) is compact and \(f\) is continuous, \(||f||_{U(I)}\) is always defined and finite.
  • The sup norm of a function is never negative. Moreover, \(||f||_{U(I)} = 0\) if and only if \(f\) is identically zero.
  • If \(c\) is a constant, then \(||cf ||_{U(I)} = |c| \, ||f||_{U(I)}\).
  • If \(f\) and \(g\) are two continuous functions on \(I\), then
    \[ ||f+g||_{U(I)} \leq ||f||_{U(I)} + ||g||_{U(I)}. \]
    Proof
    \(|f(x)| \leq \sup_{t \in I} |f(t)|\) for all \(x \in I\) and similarly for \(|g(x)|\). Thus \(\sup_{t \in I} |f(t)| + \sup_{t \in I} |g(t)|\) is an upper bound for \(|f(x)| + |g(x)|\) for all \(x \in I\). By the triangle inequality, it's also an upper bound for \(|f(x) + g(x)|\) for each \(x \in I\). Therefore
    \[{}\sup_{x \in I} |f(x) + g(x)|{}\]
    \[{}\leq \sup_{t \in I} |f(t)|{}\]
    \[{}+ \sup_{t \in I} |g(t)|. {}\]
    This is exactly the asserted inequality.
  • For a sequence \(\{f_n\}_{n=1}^\infty\) and a function \(f\), \(f_n\) converges uniformly to \(f\) as \(n \rightarrow \infty\) if and only if \(||f_n - f||_{U(I)} \rightarrow 0\) as \(n \rightarrow \infty\).
    Proof
    Uniform convergence implies that for every \(\epsilon > 0\), there is some \(N\) such that \(|f_n(x) - f(x)| < \epsilon/2\) for all \(x \in I\). This means that for these same \(n\), \(\sup_{x \in I} |f_n(x) - f(x)| \leq \epsilon/2 < \epsilon\), meaning \(||f_n - f||_{U(I)}< \epsilon\) for all \(n > N\). Conversely, if \(||f_n - f||_{U(I)} \rightarrow 0\), then there is some \(N\) such that \(n > N\) implies \(||f_n - f||_{U(I)} < \epsilon\). But then for any \(x \in I\), \(|f_n(x) - f(x)| \leq ||f_n - f||_{U(I)} < \epsilon\).

4. Uniform Cauchy Criterion

Theorem
A sequence of functions \(\{f_n\}_{n=1}^\infty\) converges uniformly on \(I\) as \(n \rightarrow \infty\) if and only if for every \(\epsilon > 0\), there is some \(N\) such that \(n,m > N\) imply that \(||f_n - f_m||_{U(I)} < \epsilon\).
Proof
\([\Rightarrow]\) Suppose is the uniform limit function. Then for every \(\epsilon\), there is some such that \(n > N\) implies . In other words, \(|f_n(x) - f(x)| < \epsilon\) for all \(x \in I\). If both \(n\) and \(m\) exceed \(N\), then the triangle inequality implies
\[{}|f_n(x) - f_m(x)|{}\]
\[{}\leq |f_n(x) - f(x)|{}\]
\[{}+ |f(x) - f_m(x)|{}\]
\[{}\leq \frac{2 \epsilon}{3}.{}\]
Since \(|f_n(x) - f_m(x)| \leq 2 \epsilon/3\) for each \(x \in I\), the supremum of this quantity over all \(x \in I\) is also at most \(2 \epsilon/3\), so it's necessarily strictly less than \(\epsilon\), i.e. \(||f_n - f_m||_{U(I)} < \epsilon\).

\([\Leftarrow]\) Being a uniform Cauchy sequence implies that \(\{f_n(x)\}_{n=1}^\infty\) is a pointwise Cauchy sequence for each \(x\), since
\[ |f_n(x) - f_m(x)| \leq ||f_n - f_m||_{U(I)}. \]
Thus \(f_n\) must converge pointwise to some limit function \(f\). Now choose \(\epsilon > 0\) and let \(N\) be a threshold such that \(n,m > N\) implies \(||f_n - f_m||_{U(I)} < \frac{\epsilon}{2}\). For each \(x \in I\), we know by pointwise convergence that all sufficiently large values of \(m\) (depending on \(x\)) have \(|f_m(x) - f(x) | < \frac{\epsilon}{2}\). By the triangle inequality and using this specific value of \(m\), every \(n > N\) has that
\[{}|f_n(x) - f(x)|{}\]
\[{}\leq |f_n(x) - f_m(x)|{}\]
\[{}+ |f_m(x) - f(x)|{}\]
\[{}< \epsilon.{}\]
In other words, all points \(x \in I\) have \(|f_n(x) - f(x)| < \epsilon\) when \(n > N\). This is exactly the definition of uniform convergence.
Example
The series \(\sum_{n=1}^\infty x^n\) converges uniformly on \([0,\frac{1}{2}]\). To prove this, use the uniform Cauchy Criterion. Let \(S_K(x)\) be the \(K\)-th partial sum, i.e., \(S_K(x) := x + x^2 + \cdots + x^K\). Then \(S_{K_1}(x) - S_{K_2}(x) = x^{K_2+1} + \cdots + x^{K_1}\) assuming that \(K_1 > K_2\). Then
\[{}||S_{K_1} - S_{k_2}||_{U([0,1/2])}{}\]
\[{}= \sup_{x \in [0,\frac{1}{2}]} (x^{K_2+1} + \cdots + x^{K_1}){}\]
\[{}= \sup_{x \in [0,\frac{1}{2}]} x^{K_2+1} (1 + \cdots + x^{K_1 - K_2-1}).{}\]
Since \(x \in [0,1/2]\) and since \(1 + \cdots + x^{K_1-K_2+1} \leq \sum_{k=0}^\infty x^k = \frac{1}{1-x}\), we have that \(||S_{K_1} - S_{K_2}||_{U([0,1/2])} \leq 2^{-K_2 - 1} \cdot 2\). So if \(N\) is sufficiently large that \(2^{-N} < \epsilon\) and if \(K_1 > K_2 > N\), then \(||S_{K_1} - S_{K_2}||_{U([0,1/2])} < \epsilon\).