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Continuity by the three-epsilon argument, the integral by the size estimate, and the derivative theorem whose hypothesis is on the derivatives.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to prove that a uniform limit of continuous functions is continuous and that the integrals converge to the integral of the limit, and to say exactly where uniformity is used in each proof. You will also be able to state the theorem for derivatives, whose hypothesis is uniform convergence of the derivatives together with convergence at one point, and to explain why the naive version is false.
Uniform convergence and the supremum distance; the size estimate for integrals; and the three failures of pointwise convergence. This lesson repairs two of the three and explains why the third needs a different hypothesis altogether.
An interchange is the claim that two limiting operations may be performed in either order: a limit with a limit, a limit with an integral, a limit with a derivative. Each theorem in this lesson is a licence for one particular interchange, and its hypothesis is the price of that licence.
Continuity passes. If each $f_n$ is continuous on $S$ and $f_n \to f$ uniformly, then $f$ is continuous.
Proof (three epsilons). For any $x, c$, $$|f(x) - f(c)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(c)| + |f_n(c) - f(c)|.$$ Choose $n$ with $\|f_n - f\|_\infty < \varepsilon/3$: that handles the outer two pieces at every point, with one index. Then that single $f_n$ is continuous at $c$, giving $\delta$ for the middle piece.
Pointwise convergence would give one index at $x$ and another at $c$, and the middle piece would be about neither.
The integral passes. If each $f_n$ is integrable on $[a,b]$ and $f_n \to f$ uniformly, then $f$ is integrable and $\int f_n \to \int f$.
Proof. $\left|\int f_n - \int f\right| = \left|\int (f_n - f)\right| \le \|f_n - f\|_\infty (b-a) \to 0$. The size estimate, applied to a gap that is small everywhere at once.
The derivative does not. $f_n(x) = \frac{\sin(nx)}{\sqrt n} \to 0$ uniformly, and $f_n'(x) = \sqrt n \cos(nx)$ converges nowhere.
The repair. If the $f_n$ are differentiable, the $f_n'$ converge uniformly, and the $f_n$ converge at one point of the interval, then the $f_n$ converge uniformly to a differentiable $f$ with $f' = \lim f_n'$. The hypothesis is about the derivatives, and the single point is needed because adding a different constant to each term leaves the derivatives untouched.
Another way: picture
Uniform convergence traps the graphs in a narrow tube about the limit. A tube controls height, and therefore controls continuity and area. It does not control slope at all: a curve can stay inside a very thin tube while oscillating steeply within it, which is exactly what the sine example does.
Another way: steps
Before exchanging a limit with something:
Integration averages and differentiation amplifies. A function that stays within $\varepsilon$ of another has an integral within $\varepsilon(b-a)$ of the other's; but its derivative can be anything at all, because a tiny wiggle of high frequency has a small height and a large slope.
That asymmetry is why the derivative theorem puts its hypothesis on the derivatives and then concludes things about the functions, rather than the other way round. Once $f_n' \to g$ uniformly, the fundamental theorem writes $f_n(x) = f_n(a) + \int_a^x f_n'$, and the integral theorem passes to the limit — giving $f(x) = f(a) + \int_a^x g$, whose derivative is $g$ by the other half of the fundamental theorem.
So the derivative theorem is the integral theorem in disguise, run through the fundamental theorem in both directions. The single point of convergence is what supplies $f(a)$.
| Object built as a uniform limit | What the theorems give |
|---|---|
| a power series inside its radius | continuity, and term-by-term integration |
| the Weierstrass nowhere-differentiable function | continuity, from a uniformly convergent series |
| the solution of a differential equation by iteration | a continuous limit, then differentiability via the integral form |
| a continuous function as a limit of polynomials | the approximation theorem of Weierstrass |
The second row is worth a moment. Weierstrass's function is $\sum 2^{-n}\cos(3^n \pi x)$: the series converges uniformly, so the sum is continuous; and it is differentiable nowhere, which was a scandal when it appeared and is now the clearest demonstration that uniform convergence preserves continuity and nothing about slopes.
The third row is how existence theorems for differential equations are proved: build a sequence by integration, get a uniform limit, and then differentiate the integral equation rather than the sequence.
Differentiating a uniform limit term by term. No theorem says this. The hypothesis must be on the derivatives.
Integrating over an unbounded interval. The estimate is the supremum distance times the width, and on an infinite interval that is useless. The spikes $f_n = 1/n$ on $[0, n]$ converge uniformly to $0$ with integrals constantly $1$.
Reading pointwise convergence into the three-epsilon proof. The proof needs one index serving both $x$ and $c$. With two indices the middle piece is about neither function.
Assuming the limit inherits differentiability because it inherits continuity. Weierstrass's function is the counterexample and it is a uniform limit of polynomials-in-disguise, each perfectly smooth.
On $[0, 1/2]$, the geometric series $\sum x^n$ converges uniformly to $\dfrac{1}{1-x}$, since the tail is bounded by $2^{-n}/(1-x) \le 2^{1-n}$.
Uniform on a closed piece inside the radius.
So the integrals may be exchanged with the sum: $\int_0^{1/2}\dfrac{dx}{1-x} = \sum_n \int_0^{1/2} x^n\,dx$.
The integral theorem, applied to the partial sums.
That is $\ln 2 = \sum_n \dfrac{(1/2)^{n+1}}{n+1}$, the series for the logarithm, obtained without differentiating anything.
A series identity, from an exchange.
Let $f_n(x) = \dfrac{\sin(nx)}{\sqrt n}$. The supremum distance to $0$ is $1/\sqrt n \to 0$: uniform on $\mathbb{R}$.
The functions converge as well as could be asked.
Their derivatives are $\sqrt n\cos(nx)$, which at $x = 0$ are $\sqrt n$.
Compute the derivatives.
So the derivatives converge at no point at all, while the functions converge uniformly everywhere.
Uniformity of the functions buys nothing here.
Pointwise limit: at $x = 0$ every term is $0$; for $0 < x \le 1$ the factor $(1-x^2)^n$ decays geometrically and beats the factor $n$, so the limit is $0$.
Find the limit first.
The integral of each term is $\dfrac{n}{2(n+1)}$, by the substitution $u = 1 - x^2$, which tends to $1/2$ rather than to $0$.
Compute the integrals exactly.
So the exchange fails, and by the theorem the convergence cannot be uniform — indeed the maximum of $f_n$ grows without bound. The theorem is being used in the contrapositive, which is often the quickest way to settle uniformity.
Build the proof that a uniform limit of continuous functions is continuous.
This task has no paper form; do it on a device.
Let $f_n(x) = x + \dfrac{1}{n}$ on the interval from $0$ to $3$. What is the limit of $\int_0^{3} f_n$?
Answer:
Match each property of the terms to whether a uniform limit inherits it.
| inherited, by splitting the gap into three pieces | inherited, together with the value of the integral | inherited, since the terms are eventually within one of the limit | not inherited: a hypothesis on the derivatives is needed instead | |
|---|---|---|---|---|
| each term is continuous | ||||
| each term is integrable on the interval from $0$ to $8$ | ||||
| each term is bounded | ||||
| each term is differentiable |
Put the steps of the proof that the integrals converge to the integral of the uniform limit, on the interval from $0$ to $7$, into order.
Number the steps in order (write the number in the box):
Here is an argument about $f_n(x) = \dfrac{\sin(nx)}{\sqrt{n}}$. Mark the one step that does not follow.
This task has no paper form; do it on a device.
Which hypotheses give $\left(\lim f_n\right)' = \lim f_n'$ on the interval from $0$ to $2$?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Build the proof that a uniform limit of continuous functions is continuous.
This task has no paper form; do it on a device.
You can run the three-epsilon proof, prove the integral theorem from the size estimate, and state the correct hypothesis for derivatives. Say in your own words why a thin tube controls area but not slope. Next: the series whose convergence is uniform on every piece inside its radius.
10. Your turn: does $\int_0^1 f_n \to \int_0^1 f$ for $f_n(x) = nx(1-x^2)^n$?, step 3