Uniform Cauchy control and differentiation of limits

Learning goals and quantifiers

Turn uniform tail control into a uniform limit, distinguish preserving continuity from preserving derivatives, and prove a sharp useful derivative theorem by the mean value theorem. For real functions on a nonempty set $A$, uniform convergence means $\sup_{x\in A}|f_n(x)-f(x)|\to0$, with the errors eventually finite. Uniform Cauchy means every $\varepsilon>0$ has $N$ such that $|f_n(x)-f_m(x)|<\varepsilon$ for all $x\in A$ and all $m,n\ge N$. If $A=\varnothing$, the epsilon–N conditions are vacuously satisfied; no real supremum of the empty set is used.

Theorem: uniform completeness

A sequence of real-valued functions is uniformly convergent if and only if it is uniformly Cauchy.

Proof. Uniform convergence gives the Cauchy estimate by two errors of size $\varepsilon/2$. Conversely, uniform Cauchy implies that for each fixed $x$, $(f_n(x))$ is a real Cauchy sequence, so define $f(x)=\lim_n f_n(x)$. Fix $\varepsilon>0$ and use uniform Cauchy with tolerance $\varepsilon/2$. For every $x$ and $n\ge N$, pass $m\to\infty$ in the estimate to get $|f_n(x)-f(x)|\le\varepsilon/2<\varepsilon$. The same $N$ works everywhere, proving uniform convergence. Completeness of the target $\mathbb R$ supplied the pointwise limits.

Derivative interchange without $C^1$ hypotheses

Theorem

Let $a<b$. Suppose every $f_n$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f_n(x_0)$ converges at one fixed $x_0\in[a,b]$ and $f_n'$ converges uniformly on $(a,b)$ to $g$, then $f_n$ converges uniformly on $[a,b]$ to a continuous $f$, differentiable on $(a,b)$, and $f'=g$. Continuity of the individual derivatives is not required.

Proof of existence. Uniform convergence of derivatives makes them uniformly Cauchy. For $m,n$ sufficiently large, the mean value theorem applied to $f_n-f_m$ between $x_0$ and any $x$ gives

$$|f_n(x)-f_m(x)|\le|f_n(x_0)-f_m(x_0)|+(b-a)\sup_{(a,b)}|f_n'-f_m'|.$$

The anchor sequence is Cauchy, so the right side is uniformly small. Uniform completeness gives a uniform limit $f$, which is continuous as a uniform limit of continuous functions.

Proof of the derivative identity. Fix an interior point $x$ and $\varepsilon>0$. Choose $N$ with $|f_n'(t)-f_N'(t)|\le\varepsilon$ for all interior $t$ and all $n\ge N$, and also $|g(x)-f_N'(x)|\le\varepsilon$. For sufficiently small nonzero $h$, apply the mean value theorem to $f_n-f_N$ on the segment from $x$ to $x+h$:

$$\left|\frac{f_n(x+h)-f_n(x)}h-\frac{f_N(x+h)-f_N(x)}h\right|\le\varepsilon.$$

Pass $n\to\infty$ at these two fixed inputs. The same inequality holds with $f$ replacing $f_n$. Since the fixed $f_N$ is differentiable at $x$, its difference quotient is within $\varepsilon$ of $f_N'(x)$ for small enough $h$. The triangle inequality bounds the difference quotient of $f$ from $g(x)$ by $3\varepsilon$. Arbitrary $\varepsilon$ proves differentiability and equality. This argument does not integrate $g$, so it does not silently assume $g$ continuous or Riemann integrable.

Worked counterexamples

Uniform functions do not suffice. The functions $f_n(x)=\sin(nx)/n$ converge uniformly to zero on every real domain, since their magnitudes are at most $1/n$. But $f_n'(0)=1$, whereas the derivative of the limit is zero. The derivative hypothesis cannot be replaced by uniform convergence of $f_n$ alone.

Pointwise derivatives do not suffice. On $[0,1]$, define $f_n(x)=\sqrt{x^2+1/n}$. These converge uniformly to $f(x)=x$, since the nonnegative difference is at most $1/\sqrt n$. Their derivatives are $x/\sqrt{x^2+1/n}$, converging pointwise to one for $x>0$ but to zero at zero. This endpoint phenomenon illustrates loss of derivative agreement at a one-sided endpoint; the theorem's differentiability conclusion is intentionally interior. For an interior failure, on $[-1,1]$ the same functions converge to $|x|$, which is not differentiable at zero, although each is smooth. The derivatives do not converge uniformly and their pointwise limit is discontinuous.

Graded practice

1. Foundation. Prove $x/n$ converges uniformly on $[-2,2]$, but not on $\mathbb R$.
2. Core. Apply the theorem to $f_n(x)=\sum_{k=1}^n x^k/2^k$ on $[-1,1]$ by controlling the derivative series.
3. Core. Show that $u_n(x)=nx(1-x^2)^n$ on $[0,1]$ tends pointwise to zero but not uniformly.
4. Proof. Explain why the convergence of $f_n$ at a single anchor cannot be omitted, even when all derivatives are identically zero.

Full solutions

1. The supremum error on $[-2,2]$ is $2/n\to0$. On the whole line it is unbounded for every $n$, so there is no uniform convergence, though there is pointwise convergence.
2. The derivative terms satisfy $|kx^{k-1}/2^k|\le k/2^k$, whose numerical series converges by the ratio test. The M-test gives uniform convergence of derivatives; all partial sums vanish at zero, providing the anchor. Thus the sum is differentiable and its derivative is the termwise derivative series. The known geometric sum is $x/(2-x)$, with derivative $2/(2-x)^2$.
3. At $x=0$ and $x=1$ the terms vanish. For fixed $0<x<1$, exponential decay of $(1-x^2)^n$ dominates $n$, so the limit is zero. At $x_n=1/\sqrt{2n}$, $u_n(x_n)=\sqrt{n/2}(1-1/(2n))^n$, and Bernoulli's inequality gives $(1-1/(2n))^n\ge1/2$. Thus $u_n(x_n)\ge\sqrt{n/2}/2\to\infty$. Thus the supremum errors do not tend to zero.
4. Take $f_n(x)=n$. Every derivative is zero and converges uniformly to zero, but the functions diverge at every point. Derivative information controls only differences of function values; the anchor fixes the additive constants.

Summary and term definitions

A uniformly Cauchy family has one tail threshold valid over the whole domain. An anchor value fixes the otherwise free additive constants. Derivative interchange requires uniform derivative control, not only small function errors. Mean-value estimates prove the theorem under weaker assumptions than the convenient continuous-derivative integral proof, and the conclusion remains explicitly local to interior points.