Uniform convergence and interchange of limits

A sequence of good functions can have a badly behaved limit. The issue is whether convergence becomes accurate at one common stage for all points. Uniform convergence supplies that control, but it does not justify every operation equally: integration is more forgiving than differentiation.

1 The order of quantifiers

A sequence $f_n:E\to\mathbb R$ converges pointwise to $f$ if for every $x\in E$ and $\varepsilon>0$ there is $N$, possibly depending on $x$, with $|f_n(x)-f(x)|<\varepsilon$ for $n\ge N$. It converges uniformly if one $N$ works for every $x\in E$. Equivalently, $\sup_{x\in E}|f_n(x)-f(x)|\to0$ when these suprema are finite.

On $[0,1]$, $f_n(x)=x^n$ tends to 0 for $x<1$ and to 1 at $x=1$. It is not uniform: for every $n$, values just below 1 make $x^n$ arbitrarily close to 1 while the limiting function is zero. The supremum error is 1, though it need not be attained.

Uniform-limit theorem. A uniform limit of continuous functions is continuous.

Proof. Fix $a$ and $\varepsilon>0$. Choose $n$ with uniform error below $\varepsilon/3$. Continuity of that one $f_n$ makes $|f_n(x)-f_n(a)|<\varepsilon/3$ for $x$ sufficiently close to $a$. The triangle inequality bounds $|f(x)-f(a)|$ by the two uniform errors and this middle term. $\square$

2 Interchanging integrals

Theorem. If $f_n$ are Riemann integrable on $[a,b]$ and converge uniformly to $f$, then $f$ is integrable and $\int f_n\to\int f$.

Proof. For any partition, $U(f,P)-L(f,P)\le U(f_n,P)-L(f_n,P)+2(b-a)\|f-f_n\|_\infty$. Choose first $n$ making the second term small, then a partition making the first term small. This proves integrability. Finally

$$\left|\int_a^b f_n-\int_a^b f\right|\le(b-a)\|f_n-f\|_\infty\to0.$$

The same proof uses rectangle volume for multiple Riemann integrals. $\square$

Pointwise convergence alone is insufficient. On $[0,1]$ define $f_n(x)=n$ for $0<x<1/n$ and zero otherwise. Every fixed point eventually has value zero, but every integral is 1. This example is integrable though not continuous; continuous triangular spikes provide the same failure.

3 Series and the uniform Cauchy criterion

A function series $\sum g_n$ converges uniformly when its partial sums do. Since $\mathbb R$ is complete, this is equivalent to: for every $\varepsilon>0$, some $N$ makes

$$\left|\sum_{k=p}^qg_k(x)\right|<\varepsilon\quad(q\ge p\ge N, x\in E).$$

Pointwise completeness supplies a candidate sum; the uniform tail bound then shows uniform convergence to it.

Weierstrass M-test. If $|g_n(x)|\le M_n$ for all $x$ and $\sum M_n<\infty$, then $\sum g_n$ converges absolutely and uniformly. The proof bounds every tail by the corresponding scalar tail and applies the criterion.

For $\sum x^n/n^2$ on $[-1,1]$, choose $M_n=1/n^2$. Its sum is continuous and the series can be integrated term by term on any subinterval. This does not alone permit termwise differentiation up to the endpoints.

4 Differentiation needs stronger control

Uniform convergence of $f_n$ alone does not imply convergence of derivatives: $f_n(x)=\sin(nx)/n$ tends uniformly to zero, but $f_n'(0)=1$ for every $n$, not the derivative of the zero limit.

Derivative theorem. Suppose $f_n\in C^1([a,b])$, $f_n(a)$ converges, and $f_n'$ converges uniformly to $g$. Then $f_n$ converges uniformly to a $C^1$ function $f$ with $f'=g$.

Proof. Uniform convergence makes $g$ continuous. Let $c=\lim f_n(a)$ and define $f(x)=c+\int_a^xg(t)dt$. The identity $f_n(x)=f_n(a)+\int_a^xf_n'$ yields uniform error at most $|f_n(a)-c|+(b-a)\|f_n'-g\|_\infty$. The fundamental theorem gives $f'=g$. $\square$

The common-value hypothesis prevents drifting constants: $f_n(x)=n$ has derivative zero but no function limit.

Exercises with complete solutions

E1 Foundation. Determine whether $x/(n+x)$ converges uniformly to zero on $[0,1]$ and on $[0,\infty)$.

S1. On $[0,1]$ the supremum is $1/(n+1)\to0$, so convergence is uniform. On the half-line the supremum is 1, approached as $x\to\infty$, so it is not uniform. Pointwise convergence still holds there.

E2 Series. Prove $\sum_{n=1}^\infty\sin(nx)/n^2$ converges uniformly on $\mathbb R$. May its derivative series be declared uniformly convergent from this argument?

S2. Bound each term by $1/n^2$ and use the M-test. Differentiating would produce $\cos(nx)/n$; at $x=0$ that series is harmonic and diverges. Therefore no uniform derivative-series claim is valid on all $\mathbb R$.

E3 Interchange. Calculate $\lim_n\int_0^1[x+x^2/n]dx$ using uniform convergence.

S3. The uniform error from $x$ is at most $1/n$, so the limit integral is $\int_0^1x dx=1/2$. Direct evaluation $1/2+1/(3n)$ verifies it.

E4 Domain dependence. Show $x^n$ converges uniformly to zero on every $[0,r]$ with $0\le r<1$, despite its failure on $[0,1]$.

S4. For $0\le r<1$, the supremum on $[0,r]$ is $r^n\to0$. The allowed stage $N$ can depend on $r$; no single estimate remains small as $r$ approaches 1.

Summary

Uniform convergence controls all domain points at one stage. It preserves continuity and permits integration on a fixed finite interval. Differentiation requires control of derivatives plus one anchored value. Always name the domain, because uniformity can disappear when it expands.