Week 6: Pointwise and Uniform Convergence

This week, we will discuss the notions of pointwise and uniform convergence.

Definition:

A sequence of functions $f_n: X \to Y$ between metric spaces is said to converge pointwise to a function $f: X \to Y$ if $f_n(x) \to f(x)$ in $Y$ for all $x \in X.$

Definition:

A sequence of functions $f_n: X \to Y$ defines a metric space $X$ with the metric of uniform convergence $d(f,g) = \sup_{x \in X}d(f(x),g(x).$ Then, a sequence $f_n$ is said to converge to $f: X \to Y$ uniformly if $f_n$ converge to $f$ in $X$ with the uniform convergence metric; that is, for every $\epsilon>0,$ there exists an $N$ such that $\sup_{x \in X} d(f(x),f_n(x)) < \epsilon$ for $n \geq N.$

Examples:

If $(y_n) \subset \mathbb{R}$ is a convergent sequence of real numbers, $y_n \to y \in \mathbb{R},$ then the constant functions $f_n: X \to \mathbb{R},$ $f_n(y) = x_n$ converge uniformly to $f(y) = x.$
The sequence $f_n:[0,1] \to \mathbb{R}$ defined by $f_n(x) = x^n$ converges pointwise to $0$ for $x<1$ and $1$ for $x=1,$ since $\lim_{n \to \infty} x^n = 0$ for $x<1$ and $\lim_{n \to \infty} 1^n = 1.$ But,
$d(f_n,0) = \sup_{x \in [0,1]} |x^n| = 1,$
so $f_n \not \to f$ uniformly.

While it is true that uniform convergence implies pointwise convergence, the preceding example shows that pointwise convergence does not necessarily imply uniform convergence.

Uniform convergence is a lot more powerful because it preserves continuity.

Proposition:

If $f_n: X \to Y$ is a sequence of continuous functions that converges uniformly to some function $f: X \to Y,$ then $f$ is continuous.

Proof:

Let $\epsilon>0.$ Then, there exists an $N$ and a $\delta>0$ such that $\sup_{x \in X} d(f_n(x),f(x)) < \frac{\epsilon}{3}$ and $d(f_n(x),f_n(y))<\frac{\epsilon}{3}$ whenever $d(x,y)<\delta$ and $n \geq N.$ Then, by the triangle inequality,

$d(f(x),f(y)) \leq d(f(x),f_n(x)) + d(f_n(x),f_n(y)) + d(f_n(y),f(y)) < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$

whenever $d(x,y) < \delta$ and $n \geq N,$ showing that $f$ is continuous.

Corollary:

For any metric space $X,$ $C(X),$ the space of continuous functions on $X,$ is closed in the space $Y$ of all functions.

The name for uniform convergence comes from the fact that it does not depend at the point of the domain that one chooses. This lets us similarly define another uniform property – uniform continuity.

Definition:

A function $f: X \to Y$ is uniformly continuous if for every $\epsilon>0$ there exists a $\delta>0$ independent of $x,y,$ such that for all $x,y \in X,$ if $d(x,y) < \delta,$ then $d(f(x),f(y))<\epsilon.$

Examples:

$f: X \to X$ given by $f(x) = x$ is uniformly continuous, since for any $\epsilon >0,$ $\delta = \epsilon$ is such that $d(x,y) < \delta$ implies $d(f(x),f(y)) = d(x,y) < \epsilon.$
$f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$ is not uniformly continuous. This can be seen as for any $\epsilon>0,$ one can find a $\delta>0$ such that $|f(x) - f(y)| = |x^2-y^2|= |x-y||x+y|<\epsilon$ whenever $|x-y|<\delta$ only if $|x+y| \leq M$ for some fixed $M >0.$ But since the domain is $\mathbb{R},$ such a $\delta$ must necessarily depend on $x,y,$ showing that $f$ is not uniformly continuous.

Proposition:

A continuous function on a compact metric space is uniformly continuous.

Proof:

Exercise.

Definition:

For a function $f: X \to \mathbb{R},$ define the supremum norm of $f$ to be

$\|f\|_\infty = d(f,0) = \sup_{x \in X} f(x).$

Theorem (Weierstrass M-Test):

Suppose $f_n : X \to Y$ is a sequence of bounded continuous functions such that $\sum_{n=1}^\infty \|f_n\|_{\infty}$ converges. Then, the function $f$ defined by $f(x) = \sum_{n=1}^\infty f_n(x)$ is continuous, and the sum converges uniformly to $f.$

Proof:

Since $\sum_{n=1}^\infty |f_n(x)| \leq \sum_{n =1}^\infty \|f_n\|_{\infty},$ and the latter sum converges, the former sum is absolutely convergent, so by the Absolute Convergence Test, $\sum_{n=1}^\infty f_n(x) = f(x)$ exists. Define $f: X \to \mathbb{R}$ pointwise by $f(x) = \sum_{n=1}^\infty f_n(x).$ It remains to show that $f$ is continuous, for which we use the fact that the limit of a uniformly convergent sequence of functions is continuous. We see by the triangle inequality that

$d\left(f,\sum_{n=1}^k f_n\right) = \sup_{x \in X} \left|\sum_{n=1}^\infty f_n(x) - \sum_{n=1}^k f_n(x) \right| = \sup_{x \in X} \left|\sum_{n=k+1}^\infty f_n(x) \right| \leq \sum_{n=k+1}^\infty \|f_n\|_\infty < \epsilon$

for sufficiently large $n$ since by assumption, the sum converges. Thus, $f_n \to f$ converges uniformly, and since each $f_n$ is continuous, $f$ is continuous.