Sequence of Continuous Functions Converges Uniformly
Uniform Convergence
[under construction]
The problems with pointwise convergence
Definition. If $X$ is a metric space, and $f_n:X\to \R$ ($n\in\N$) is a sequence of functions, then $f_n$ converges pointwise to $f$ if for every $x\in X$ one has $\lim_{n\to\infty} f_n(x) = f(x)$.
For example, the sequence of functions $f_n(x) = x^n/n$ converges pointwise to zero on the interval $X=[-1,1]$, because for each $x\in [-1,1]$ one has $|x^n/n|\leq 1/n$, and thus \[ \lim_{n\to\infty} \frac{x^n}{n} = 0. \]
The limit of a pointwise convergent sequence of continuous functions does not have to be continuous. For example, consider $X=[0,1]$, and $f_n(x) = x^n$. Then \[ \lim_{n\to\infty} f_n(x) = f(x) = \begin{cases} 0 & (0\leq x\lt 1) \\ 1 & (x=1) \end{cases} \]
The derivatives of a pointwise convergent sequence of functions do not have to converge. $X=\R$, $f_n(x) = \frac1n \sin(n^2x)$. In this case \[ \lim_{n\to\infty} f_n(x) = 0, \] so the pointwise limit function is $f(x) = 0$: the sequence of functions converges to $0$. What about the derivatives of the sequence? These are given by \[ f_n'(x) = n\cos (n^2x), \] and for most $x\in\R$ the sequence $n \cos(n^2x)$ is unbounded. The sequence of derivatives $f_n'(x)$ does not converge pointwise.
The integrals of a pointwise convergent sequence of functions do not have to converge. Consider $X=[0,1]$, $f_n(x) = \frac{2n^2x}{\bigl(1+n^2x^2\bigr)^2}$. Then \[ \lim_{n\to\infty} f_n(x) = 0 \] for all $x\in[0,1]$. But the integrals of $f_n$ over the interval $X$ are \[ \int_0^1 \frac{2n^2xdx}{\bigl(1+n^2x^2\bigr)^2} \stackrel{u=1+n^2x^2}= \int_1^{1+n^2} \frac{du}{u^2} = 1 - \frac1{1+n^2}. \] Therefore, even though $\lim_{n\to\infty} f_n(x) = 0$ for all $x\in [0,1]$, we have \[ \lim_{n\to\infty} \int_0^1 f_n(x) dx = 1. \]
Uniform convergence
Definition.A sequence of functions $f_n:X\to Y$ converges uniformly if for every $\epsilon\gt0$ there is an $N_\epsilon\in\N$ such that for all $n\geq N_\epsilon$ and all $x\in X$ one has $d(f_n(x), f(x))\lt \epsilon$.
Uniform convergence implies pointwise convergence, but not the other way around. For example, the sequence $f_n(x) = x^n$ from the previous example converges pointwise on the interval $[0,1]$, but it does not converge uniformly on this interval. To prove this we show that the assumption that $f_n(x)$ converges uniformly leads to a contradiction.
If $f_n(x)$ converges uniformly, then the limit function must be $f(x) = 0$ for $x\in[0,1)$ and $f(1) = 1$. Uniform convergence implies that for any $\epsilon\gt0$ there is an $N_\epsilon\in\N$ such that $|x^n- f(x)|\lt \epsilon$ for all $n\geq N_\epsilon$ and all $x\in [0,1]$. Assuming this is indeed true we may choose $\epsilon$, in particular, we can choose $\epsilon=\frac12$. Then there is an $N\in \N$ such that for all $n\geq N$ we have $|x^n-f(x)|\lt \frac 12$. We may choose $n$ and $x$. Let us choose $n=N$, and $x=\bigl(\frac34\bigr)^{1/N}$. Then we have $f(x)=0$ and thus \[ |f_N(x) - f(x)| = x^N - 0 = \frac 34 \gt \frac12, \] contradicting our assumption.
The uniform metric
Definition—the set of bounded functions. If $E$ is a set, then a function $f:E\to\R$ is bounded if there is an $M\in\R$ such that $|f(x)|\leq M$ for all $x\in E$. We will write $\cB(E)$ for the set of all bounded functions from $E$ to $\R$.
Definition—the uniform distance between bounded functions. The uniform distance between two bounded functions $f, g\in\cB(E)$ is \[ \du(f, g) = \sup_{x\in E} |f(x) - g(x)|. \]
The uniform distance $d(f,g)$ between two functions $f$ and $g$ is the least upper bound of all possible values that $|f(x)-g(x)|$ can have if $x\in E$.
Theorem. $\du(f, g)$ is a metric on $\cB(E)$.
We have to check:
- $\du(f,g) = \du(g,f)$
- $\du(f,g)\geq 0$ with $\du(f,g) = 0$ if and only if $f=g$
- $\du(f,g) \leq \du(f,h) + \du(h,g)$
Theorem. The uniform metric $\du$ is a metric on $\cB(E)$. A sequence of bounded functions $f_n:E\to\R$ converges uniformly to $f$ if and only if $\du(f_n, f)\to 0$, i.e. if and only if $f_n$ converges to $f$ in the sense of convergence in the metric space $\cB(E)$.
Example. Let $E = [0,1)$ and consider the sequence of functions $f_n(x) = x^n$. We know that $f_n(x)\to0$ pointwise on $[0,1)$.
Question:Does the sequence converge uniformly on $[0,1)$?
Answer: Since uniform convergence is equivalent to convergence in the uniform metric, we can answer this question by computing $\du(f_n, f)$ and checking if $\du(f_n, f)\to0$. We have, by definition \[ \du(f_n, f) = \sup_{0\leq x\lt 1}|x^n - 0| =\sup_{0\leq x\lt 1} x^n = 1. \] Therefore \[ \lim_{n\to\infty} \du(x^n, 0) = \lim_{n\to\infty} 1 = 1 \neq 0. \] The sequence of functions $x^n$ does not converge uniformly on the interval $[0,1)$.
Left: the uniform distance between $f_n(x)= x^n$ and $f(x)=0$ on the interval $0\leq x\lt 1$ is $d(f_n, f)=1$, so, on the interval $[0,1)$, $x^n$ does not converge uniformly to $0$ as $n\to\infty$. Right: on the interval $[0,a]$ the uniform distance between $f_n(x)= x^n$ and $f(x) = 0$ is $a^n$. Since $a\lt1$, we have $a^n\to0$ so that $f_n(x)=x^n$ does converge uniformly to zero on the interval $[0,a]$.
Question: Does the same sequence of functions converge uniformly on the interval $[0,a]$ if $a$ is some number with $0\lt a \lt 1$?
Answer: We again compute the uniform distance between the functions $f_n(x) = x^n$ and $f(x)=0$, but this time on the interval $[0,a]$ instead of the interval $(0,1)$ that we used in the previous example. We have \[ \du(x^n, 0) = \sup_{0\leq x\leq a} |x^n-0| = \sup_{0\leq x\leq a} x^n = a^n. \] Since $0\lt a\lt 1 $ we have \[ \lim_{n\to\infty} \du(f_n, f) = \lim_{n\to\infty} a^n = 0, \] so that the sequence of functions $x^n$ does converge uniformly on the interval $[0,a]$, for any $a\in(0,1)$.
Three consequences of uniform convergence
In the following theorems $E$ is a metric space.Theorem. Let $f_n:E\to \R$ be a sequence of functions. If $f_n$ converges uniformly to $f:E\to \R$, and if each $f_n$ is continuous, then $f$ is also continuous.
This theorem implies that for any uniformly convergent sequence of functions $f_n:E\to \R$ and convergent sequence of points $x_n\in E$ one can "switch limits," i.e. \[ \lim_{n\to\infty} \lim_{k\to\infty} f_n(x_k) = \lim_{k\to\infty} \lim_{n\to\infty} f_n(x_k) . \]
Theorem.
If $f_n:[a,b]\to\R$ is a sequence of Riemann integrable functions that converges uniformly to $f:[a,b]\to\R$, then the limit $f$ is also Riemann integrable and \[ \lim_{n\to\infty} \int_a^b f_n(x) dx = \int_a^b f(x) dx. \]
Since $f(x) = \lim_{n\to\infty} f_n(x)$ we can write the conclusion as \[ \lim_{n\to\infty} \int_a^b f_n(x) dx = \int_a^b \lim_{n\to\infty} f_n(x) dx, \] In other words, this theorem justifies switching limits and integration.
Theorem.
Let $f_n:[a,b]\to\R$ be a sequence of differentiable functions whose derivatives $f_n'$ are continuous. If $f_n$ converges uniformly to $f$ and $f_n'$ converges uniformly to $g$, then the limit $f$ is differentiable and its derivative is $f'=g$.
We can rewrite the conclusion as \[ \lim_{n\to\infty}\frac{d f_n(x)}{dx} = \frac{d \lim_{n\to\infty}f_n(x)}{dx}. \] This theorem justifies switching of limits and derivatives. Note that the hypotheses of the theorem require that both the sequence of functions $f_n$ and the sequence of their derivatives $f_n'$ must converge uniformly.
Uniform convergence of series
Pointwise convergence for series.If $f_n$ is a sequence of functions defined on some set $E$, then we can consider the partial sums \[ s_n(x) = f_1(x) + \cdots + f_n(x) = \sum_{k=1}^n f_k(x). \] If these converge as $n\to\infty$, and if this happens for every $x\in E$, then we say that the series converges pointwise. The sum of the series is \[ S(x) = \sum_{k=1}^\infty f_k(x) \stackrel{\sf def}= \lim_{n\to\infty}\sum_{k=1}^n f_k(x) = \lim_{n\to\infty} s_n(x). \] The sum $\sum_1^\infty f_k(x)$ is defined for each $x\in E$ and so it is a function on $E$.
Uniform convergence of series. A series $\sum_{k=1}^\infty f_k(x)$ converges uniformly if the sequence of partial sums $s_n(x) = \sum_{k=1}^n f_k(x)$ converges uniformly.
The Weierstrass M–test. If $f_n:E\to\R$ is a sequence of functions for which one has a sequence $M_n$ with \[ |f_n(x)|\leq M_n \text{ for all }x\in E, \] and for which \[ \sum_{n=1}^\infty M_n \lt \infty, \] then the series $\sum_{k=1}^\infty f_k(x)$ converges uniformly.
How to prove a series does not converge uniformly. If $\sum f_n(x)$ is a series whose terms $f_n$ are functions on some set $E$, and if the series converges uniformly on $E$, then
$\displaystyle\lim_{n\to\infty} \Bigl(\sup_{x\in E} |f_n(x)|\Bigr) =0$.
Let $s_n(x) = f_1(x) + \cdots + f_n(x)$ be the $n$th partial sum of the series, and let $S(x) = \sum_1^\infty f_n(x)$ be the sum of the series. Since the series converges uniformly, we have $\lim_{n\to\infty} \du(s_n, S) = 0$. By the triangle inequality we also have
$\du(s_n, s_{n-1}) \leq \du(s_{n-1}, S) + \du(S, s_n)$
so that $\du(s_n, s_{n-1}) \to 0$ as $n\to\infty$.
The uniform distance between the two consecutive partial sums $s_{n-1}$ and $s_n$ is
$\du(s_{n-1}, s_n) = \sup_{x\in E} |s_{n-1}(x) - s_n(x)| = \sup_{x\in E} |f_n(x)|$.
It follows that
$\sup_{x\in E} |f_n(x)| \to 0$ as $n\to\infty$.
Source: https://people.math.wisc.edu/~angenent/521.2017s/UniformConvergence.html
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