3.2 Uniform convergence and interchange of limits

Proof

Prerequisite proofs used below

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Uniform limits of proper Riemann integrable real functions

Let $a<b$ be finite and let each real $f_n$ be properly Riemann integrable on $[a,b]$. Suppose $f_n\to f$ uniformly there. A single sufficiently late $f_N$ is bounded and has bounded uniform error, so $f$ is bounded. Put $L=b-a>0$. Given $\varepsilon>0$, choose $N$ with $\eta=\sup|f-f_N|<\varepsilon/(4L)$. For any partition cell $I$, the supremum of $f$ is at most the supremum of $f_N$ plus $\eta$, and its infimum is at least the infimum of $f_N$ minus $\eta$. Hence

$$U(f,P)-L(f,P)\le U(f_N,P)-L(f_N,P)+2\eta(b-a).$$

Choose $P$ for the fixed integrable $f_N$ with its Darboux gap less than $\varepsilon/2$. The displayed gap for $f$ is then less than $\varepsilon$, so the actually proved Darboux criterion gives integrability of $f$. Only after this step do we subtract their proper integrals. The bounds for an integrable function and $|f-f_n|\le\sup|f-f_n|$ give

$$\left|\int_a^b f-\int_a^b f_n\right|\le (b-a)\sup_{[a,b]}|f-f_n|\longrightarrow0.$$

For $a=b$, all these oriented integrals are zero and no division by $L=0$ is used; reversed endpoints follow by changing signs. No assertion about improper integrals or noncompact domains follows from this theorem.

Pointwise versus uniform

Definition

Two modes of convergence

For functions $f_n,f$ on $A$, pointwise convergence means: for every $x\in A$ and every $\varepsilon>0$, an $N$ exists such that $n\ge N$ implies $|f_n(x)-f(x)|<\varepsilon$; $N$ may depend on $x$. Uniform convergence means: for every $\varepsilon>0$ there is one $N$ such that for every $n\ge N$ and every $x\in A$, $|f_n(x)-f(x)|<\varepsilon$.

For nonempty $A$, equivalently, uniform convergence means $\sup_{x\in A}|f_n(x)-f(x)|\to0$, with the suprema eventually finite. To disprove it, find $\varepsilon_0>0$, indices $n_k\to\infty$, and points $x_k\in A$ with $|f_{n_k}(x_k)-f(x_k)|\ge\varepsilon_0$.

Worked example

Pointwise convergence can lose continuity

On $[0,1]$, $f_n(x)=x^n$ converges pointwise to $f(x)=0$ for $0\le x<1$ and $f(1)=1$. Each $f_n$ is continuous but the limit is discontinuous at $1$. The convergence cannot be uniform; for example $x_n=1-1/n$ gives $x_n^n\to e^{-1}$ while $f(x_n)=0$. On $[0,r]$ for $0<r<1$, $\sup|x^n|=r^n\to0$, so convergence is uniform there.

Theorem

Uniform limit preserves continuity

If each $f_n$ is continuous at $a\in A$ relative to $A$ and $f_n\to f$ uniformly on $A$, then $f$ is continuous at $a$.

Proof

The three-term estimate

Given $\varepsilon>0$, choose $N$ so $|f_N(x)-f(x)|<\varepsilon/3$ for all $x\in A$. By continuity of the fixed function $f_N$ at $a$, choose $\delta$ so $|f_N(x)-f_N(a)|<\varepsilon/3$ whenever $|x-a|<\delta$. Then

$$|f(x)-f(a)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(a)|+|f_N(a)-f(a)|<\varepsilon.$$

Integration, differentiation and series

If Riemann-integrable $f_n$ converge uniformly to $f$ on $[a,b]$, then $f$ is integrable and $\int_a^b f_n\to\int_a^b f$. Indeed the integral difference is bounded by $(b-a)\sup|f_n-f|$; integrability of the uniform limit follows by combining a sufficiently small uniform error with a partition for a fixed $f_N$.

Uniform convergence of $f_n$ alone does not guarantee convergence of derivatives. For example $f_n(x)=\sin(nx)/n$ converges uniformly to zero on $\mathbb R$, but $f_n'(0)=1$ for every $n$, whereas the limit's derivative is zero. A useful sufficient theorem is: if $a<b$, $f_n$ are $C^1$ on $[a,b]$, $f_n(x_0)$ converges for one $x_0\in[a,b]$, and $f_n'$ converges uniformly to $g$, then $f_n$ converges uniformly to $f$, differentiable on $(a,b)$ with $f'=g$ there. The fundamental theorem and the uniform-integration result give $f(x)=\lim f_n(x_0)+\int_{x_0}^xg(t)dt$; continuity of $g$ follows from uniform convergence of continuous derivatives.

For a series of functions, uniform convergence means uniform convergence of its partial sums. The Weierstrass M-test says that if $|u_n(x)|\le M_n$ for all $x\in A$ and $\sum M_n$ converges, then $\sum u_n$ converges uniformly and absolutely at each point. Its proof uses the uniform Cauchy tail bound $|\sum_{k=m}^n u_k(x)|\le\sum_{k=m}^nM_k$.

Worked example

A uniform series bound

For $u_n(x)=x^n/2^n$ on $[0,1]$, $|u_n(x)|\le2^{-n}$. The geometric majorant converges, so the M-test gives uniform convergence. Each partial sum is continuous, so the limit is continuous. In fact its sum is $x/(2-x)$ by the geometric formula.

Method comparison

For pointwise convergence fix $x$ first. For uniform convergence seek a supremum bound or a summable majorant independent of $x$. For failure choose a moving point $x_n$ that exposes the worst behavior. Interchanging differentiation requires more than uniform convergence of functions; write the exact sufficient hypotheses before using the theorem.

Practice

1. Does $f_n(x)=x/n$ converge uniformly to zero on $[0,1]$? On $\mathbb R$?
2. Show $x^n\to0$ uniformly on $[0,1/2]$.
3. Use the M-test for $\sum_{n=1}^\infty x^n/3^n$ on $[-1,1]$.
4. Give an example of uniformly convergent differentiable functions whose derivatives do not converge to the derivative of the limit at a point.

Full solutions

1. On $[0,1]$, $\sup|x/n|=1/n\to0$, so yes. On $\mathbb R$, the error is unbounded for each $n$, so no uniform convergence, although convergence is pointwise.
2. $\sup_{[0,1/2]}|x^n|=(1/2)^n\to0$.
3. $|x^n/3^n|\le3^{-n}$ and the geometric majorant converges, so the series converges uniformly and absolutely pointwise.
4. $f_n(x)=\sin(nx)/n$ converges uniformly to zero, but $f_n'(0)=1$ for all $n$, not $0$.

Synthesis

Uniform convergence places one tail bound across the whole domain. It preserves continuity and supports integration on a compact interval; differentiation needs an additional theorem and hypotheses.