MAT159H1/Unit 06
Unit 06 · Analysis 1B

Uniform Convergence

The capstone idea of the course: a stronger kind of convergence for sequences and series of functions, strong enough to let you swap limits with integrals and derivatives. Power series, next unit, are the main example.

Why it matters & where it comes from
Why it exists

Pointwise convergence is too weak: a limit of continuous functions can be discontinuous, and you cannot freely swap limits with integrals or derivatives. Uniform convergence is the extra condition that fixes both problems at once.

In the real world

It justifies term-by-term differentiation and integration of series — the silent step behind Fourier analysis, differential-equation solutions, and much of applied mathematics and numerical computing.

Story & open questions

Early 19th-century mathematicians, including Cauchy, made famous errors by assuming limits of continuous functions were always continuous. Weierstrass isolated uniform convergence in the 1840s–50s as the missing hypothesis — one of the pivotal moments in the rigorisation of analysis.

By the end you can

Two kinds of convergence

Definition

$f_n\to f$ pointwise on $S$ if $f_n(x)\to f(x)$ for each fixed $x\in S$ (the $N$ in the ε–N definition may depend on $x$). $f_n\to f$ uniformly on $S$ if $\sup_{x\in S}|f_n(x)-f(x)|\to0$ — one $N$ works for every $x\in S$ simultaneously.

Example Pointwise but not uniform

$f_n(x)=x^n$ on $[0,1]$.

Pointwise limit: $f(x)=0$ for $x\in[0,1)$, $f(1)=1$ — a discontinuous function, even though every $f_n$ is continuous.
$\sup_{x\in[0,1]}|x^n-f(x)|=1$ for every $n$ (approached as $x\to1^-$), so the convergence is not uniform.

The Weierstrass M-test

Theorem M-test

Suppose $|f_n(x)|\le M_n$ for all $x\in S$, with $\sum M_n$ convergent. Then $\sum f_n$ converges uniformly (and absolutely) on $S$. The idea: bounding each term by a number independent of $x$ turns a question about functions into an ordinary series question.

Example

$\displaystyle\sum_{n=1}^\infty\frac{\cos(nx)}{n^2}$ converges uniformly on all of $\mathbb{R}$: $\left|\dfrac{\cos nx}{n^2}\right|\le\dfrac1{n^2}=M_n$, and $\sum1/n^2$ converges.

Why uniform convergence matters

Theorem Continuity is preserved

If $f_n\to f$ uniformly on $S$ and each $f_n$ is continuous, then $f$ is continuous.

Proof sketch
Fix $x_0\in S$ and $\varepsilon>0$. Uniform convergence gives $N$ with $|f_N(x)-f(x)|<\varepsilon/3$ for every $x\in S$ at once.
Continuity of $f_N$ at $x_0$ gives $\delta>0$ with $|x-x_0|<\delta\Rightarrow|f_N(x)-f_N(x_0)|<\varepsilon/3$.
Combine (the "add and subtract" trick again): $|f(x)-f(x_0)|\le|f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|< \varepsilon$. ∎ Note the middle term is the only one using continuity of a single $f_N$; the outer two rely entirely on uniformity.
Swapping integrals and derivatives

If $f_n\to f$ uniformly on $[a,b]$ and each $f_n$ is integrable, then $\int_a^b f=\lim_n\int_a^b f_n$. For derivatives: if $f_n\to f$ pointwise at one point, each $f_n'$ exists and is continuous, and $f_n'\to g$ uniformly, then $f$ is differentiable with $f'=g$ — note the stronger hypothesis needed on the derivatives, not the functions themselves.

Why derivatives need more care Uniform convergence of $f_n$ alone does not give convergence of $f_n'$: $f_n(x)=\tfrac1n\sin(n^2x)$ converges uniformly to $0$, but $f_n'(x)=n\cos(n^2x)$ does not converge at all. This is why the derivative theorem hypothesizes uniform convergence of $f_n'$ directly, not of $f_n$.
Common pitfalls

• Pointwise convergence does not preserve continuity and does not allow swapping limits with integrals or derivatives in general.
• Always compute the $\sup$ of the difference over the whole domain, not just check a few points.
• The M-test only ever proves uniform convergence — it can never be used to show a series does not converge uniformly.
• Uniform convergence of $f_n$ says nothing about $f_n'$ by itself; the derivative-swap theorem needs uniform convergence of the derivatives.

Practice

Practice 1

Show $f_n(x)=x/n$ converges uniformly to $0$ on $[0,1]$.

Answer
$\sup_{[0,1]}|x/n|= 1/n\to0$.
Practice 2

Show $f_n(x)=x^n$ is not uniform on $[0,1]$.

Answer
$\sup|x^n-f(x)|=1$ (approached as $x\to1^-$) for every $n$, so it does not tend to $0$.
Practice 3

Does $f_n(x)=\dfrac{\sin(nx)}{n}$ converge uniformly on $\mathbb{R}$?

Answer
Yes; $\sup|f_n(x)-0|\le1/n\to0$.
Practice 4 · M-test

Show $\displaystyle\sum_{n=1}^\infty\dfrac{x^n}{n^2}$ converges uniformly on $[-1,1]$.

Answer
$\left|\tfrac{x^n} {n^2}\right|\le\tfrac1{n^2}=M_n$ on $[-1,1]$, and $\sum1/n^2$ converges, so the M-test applies.
Practice 5 · Not uniform on an open interval

Show $\sum x^n$ converges uniformly on $[-\tfrac12,\tfrac12]$ but not on $(-1,1)$.

Solution
On $[-\tfrac12,\tfrac12]$: $|x^n|\le2^{-n}=M_n$ and $\sum2^{-n}$ converges, so M-test applies. On $(-1,1)$ no single convergent $\sum M_n$ bounds $|x^n|$ uniformly, since $\sup_{(-1,1)}|x|^n=1$ for every $n$.
Practice 6 · Swapping integrals

Given $f_n\to f$ uniformly on $[0,1]$ with each $f_n$ continuous, explain why $\lim_n\int_0^1 f_n(x)\,dx=\int_0^1 f(x)\,dx$.

Answer
$\left|\int_0^1f_n-\int_0^1f\right|\le\int_0^1|f_n-f|\le \sup_{[0,1]}|f_n-f|\to0$, using uniform convergence to bound the integral of the difference by the sup-norm times the interval length.
Practice 7 · The derivative trap

Verify $f_n(x)=\tfrac1n\sin(n^2x)\to0$ uniformly on $\mathbb{R}$, but $f_n'(0)$ does not converge.

Answer
$\sup|f_n(x)|\le1/n\to 0$, so $f_n\to0$ uniformly. But $f_n'(x)=n\cos(n^2x)$, so $f_n'(0)=n\to\infty$ — it does not converge at all, showing uniform convergence of $f_n$ says nothing about $f_n'$.
Practice 8 · Continuity contrapositive

$f_n(x)=x^n$ on $[0,1]$ has a discontinuous pointwise limit. What does this immediately tell you about the mode of convergence, without any further computation?

Answer
By the contrapositive of "uniform limit of continuous functions is continuous," since each $f_n$ is continuous but the limit is not, the convergence cannot be uniform.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT159H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT159H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The uniform-convergence spine is: identify the pointwise limit, estimate the worst-case error, then use uniform convergence to preserve structure.

Study routine. Never interchange a limit with an integral or derivative until the relevant uniform hypothesis has been checked.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Pointwise not uniform

Analyze $f_n(x)=x^n$ on $[0,1]$.

Guide
The pointwise limit is $0$ on $[0,1)$ and $1$ at $1$, so convergence is not uniform because uniform limits of continuous functions are continuous.
Extension 2 · Sup norm

Show $f_n(x)=x/n$ converges uniformly to $0$ on $[0,1]$.

Guide
The sup error is $\sup_{[0,1]} x/n=1/n\to0$.
Extension 3 · M-test

Prove $\sum x^n/n^2$ converges uniformly on $[-1,1]$.

Guide
Since $|x^n/n^2|\le1/n^2$ and $\sum1/n^2$ converges, the M-test applies.
Extension 4 · Continuity preservation

If continuous $f_n$ converge uniformly to $f$, what can you conclude?

Guide
The limit function $f$ is continuous. The proof uses a three-term epsilon argument with uniform convergence and continuity of one chosen $f_N$.
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