Uniform convergence of function sequences: Worked examples

The scenario matrix

Every uniform-convergence exercise falls into one of these cells. The examples below are tagged with the cell(s) they cover.

Cell	Situation	What can go wrong / what to watch
A	Limit is continuous, convergence uniform	benign — bound $M_{n}$ by a formula independent of $x$
B	Limit is discontinuous ⇒ never uniform	jump is the fingerprint of failure
C	A travelling bump (max sneaks across the domain)	$M_{n}$ can stay constant even if $f \equiv 0$
D	Uniformity depends on the domain (fails on $E$ , holds on $E^{'} \subset E$ )	shrink away from the trouble point
E	Degenerate / zero input: $x = 0$ , endpoints, $f_{n}$ already $= f$	check that trivial points don't break the sup
F	Limiting behaviour: sup attained by calculus (find max with a derivative)	locate where the peak lives at step $n$
G	Series case — use Cauchy / Weierstrass M-test without knowing $f$	bound each term by a summable constant
H	Word problem / real-world framing (an "error budget" deadline)	translate "for all users at once" = uniform
I	Exam twist: convergence uniform but derivatives/integrals misbehave	uniform of $f_{n}$ ≠ uniform of $f_{n}^{'}$

Six figures below carry the geometry of cells B, C, D, F.

Example 1 — Benign uniform (Cell A)

Worked example

$f_{n} (x) = \frac{x}{n}$ on $[0, 1]$ Forecast: Guess now — does the whole graph flatten to $0$ at one shared pace, or does some point lag behind?

Step 1. Find the pointwise limit. Why this step? We cannot measure a gap without knowing $f$ . For each fixed $x$ , $\frac{x}{n} \to 0$ as $n \to \infty$ , so $f (x) = 0$ .

Step 2. Write the gap. Why this step? $M_{n}$ is the sup of $∣ f_{n} - f ∣$ . Here $∣ f_{n} (x) - f (x) ∣ = \frac{x}{n} - 0 = \frac{x}{n}$ .

Step 3. Take the sup over $x \in [0, 1]$ . Why this step? $\frac{x}{n}$ increases in $x$ , so the largest value is at the right endpoint $x = 1$ : $M_{n} = sup_{x \in [0, 1]} \frac{x}{n} = \frac{1}{n} .$

Step 4. Test $M_{n} \to 0$ . $\frac{1}{n} \to 0$ , so convergence is uniform. ✅

Verify: The bound $\frac{1}{n}$ does not mention $x$ — the hallmark of uniformity. Sanity check a value: at $n = 100$ , no point deviates more than $0.01$ ; the whole graph is inside a horizontal band of half-width $0.01$ .

Example 2 — Discontinuous limit kills uniformity (Cell B)

Worked example

$f_{n} (x) = \frac{1}{1 + x ^{2 n}}$ on $[0, 2]$ Forecast: Each $f_{n}$ is a smooth bump. Guess the limit shape — does it stay smooth?

Step 1. Pointwise limit — split by size of $x$ . Why this step? $x^{2 n}$ behaves completely differently for $x < 1$ , $x = 1$ , $x > 1$ , so we must case-split.

$0 \leq x < 1$ : $x^{2 n} \to 0$ , so $f_{n} \to \frac{1}{1 + 0} = 1$ .
$x = 1$ : $x^{2 n} = 1$ , so $f_{n} = \frac{1}{2}$ for all $n$ ; limit $= \frac{1}{2}$ .
$1 < x \leq 2$ : $x^{2 n} \to \infty$ , so $f_{n} \to 0$ .

So $f (x) = ⎩ ⎨ ⎧ 1 \frac{1}{2} 0 0 \leq x < 1 x = 1 1 < x \leq 2$ — a jump at $x = 1$ .

Step 2. Invoke the continuity theorem contrapositive. Why this step? Every $f_{n}$ is continuous. If convergence were uniform, the limit would be continuous. It is not — it jumps at $x = 1$ . Therefore convergence is not uniform. ❌

Step 3. Confirm with $M_{n}$ directly. Why this step? We like a second, computational witness. Approach $x \to 1^{-}$ : there $f (x) = 1$ but $f_{n} (x) = \frac{1}{1 + x ^{2 n}}$ dips toward $\frac{1}{2}$ . At exactly $x = 1$ , $∣ f_{n} (1) - f (1) ∣ = ∣ \frac{1}{2} - \frac{1}{2} ∣ = 0$ , but just left of $1$ the gap climbs to $\frac{1}{2}$ . So $M_{n} \geq \frac{1}{2}$ for all $n$ , hence $M_{n} \neq \to 0$ .

Verify: Two independent arguments (discontinuity + sup) agree. The look (figure): the curves stay near $1$ then plunge, and the plunge never sharpens into a step you can trap in a thin band.

Example 3 — The travelling bump (Cell C, F)

Worked example

$f_{n} (x) = n x e^{- n x}$ on $[0, \infty)$ Forecast: For each fixed $x > 0$ this $\to 0$ (exponential beats the factor $n$ ). So $f \equiv 0$ . Does that mean uniform? Guess before reading on.

Step 1. Pointwise limit. Why this step? At $x = 0$ , $f_{n} (0) = 0$ . For fixed $x > 0$ , $e^{- n x}$ decays exponentially while $n x$ grows only linearly, so $n x e^{- n x} \to 0$ . Thus $f (x) = 0$ everywhere.

Step 2. Locate the peak of $f_{n}$ using a derivative. Why this step? $M_{n} = sup_{x} n x e^{- n x}$ ; to find where the graph is tallest we set the derivative to zero — that is the calculus tool that answers "where is the maximum?" $\frac{d}{d x} (n x e^{- n x}) = n e^{- n x} (1 - n x) = 0 ⟹ x_{n} = \frac{1}{n} .$ The peak lives at $x_{n} = \frac{1}{n}$ , which slides toward $0$ as $n$ grows — a moving bump.

Step 3. Height of the peak. Why this step? Plug $x_{n} = \frac{1}{n}$ back in: $M_{n} = n \cdot \frac{1}{n} \cdot e^{- n \cdot \frac{1}{n}} = e^{- 1} \approx 0.368.$

Step 4. Test. $M_{n} = e^{- 1} \neq \to 0$ . Not uniform on $[0, \infty)$ . ❌

Verify: Look at the figure — every curve has the same height $1/ e$ , just its peak marching left. No single horizontal band of height $< 1/ e$ can ever contain all the tails. Numerically $e^{- 1} = 0.3679$ , constant in $n$ .

Example 4 — Uniformity depends on the domain (Cell D)

Worked example Same

$f_{n} (x) = n x e^{- n x}$ , now on $[δ, \infty)$ with fixed $δ > 0$ Forecast: We just failed on $[0, \infty)$ because the bump reaches $x = 0$ . If we forbid $x$ near $0$ , can we win?

Step 1. Where is the peak relative to $δ$ ? Why this step? The unconstrained peak is at $x_{n} = \frac{1}{n}$ . Once $n > \frac{1}{δ}$ (i.e. $\frac{1}{n} < δ$ ), the peak has slid out of the domain — it sits to the left of $δ$ . On $[δ, \infty)$ the function $n x e^{- n x}$ is then decreasing, so its sup is at the left endpoint $x = δ$ .

Step 2. Evaluate the sup. Why this step? For $n > \frac{1}{δ}$ , $M_{n} = n δ e^{- n δ} .$

Step 3. Take the limit. Why this step? $e^{- n δ}$ decays exponentially in $n$ and crushes the linear $n δ$ , so $n δ e^{- n δ} \to 0$ . Uniform on $[δ, \infty)$ . ✅

Verify: Same formula, opposite verdict — exactly the moral from the parent's $x^{n}$ examples: uniformity is a property of the (function, domain) pair. Numerically with $δ = 1$ : $M_{5} = 5 e^{- 5} \approx 0.0337$ , $M_{10} = 10 e^{- 10} \approx 4.54 \times 1 0^{- 4}$ — clearly heading to $0$ .

Example 5 — Degenerate / already-equal case (Cell E)

Worked example

$f_{n} (x) = \frac{x ^{2}}{x ^{2} + \frac{1}{n}}$ on $R$ , with special attention to $x = 0$ Forecast: At $x = 0$ the formula gives $\frac{0}{0 + 1/ n} = 0$ for every $n$ — a point where $f_{n}$ never moves. Does that lonely point cause trouble?

Step 1. Pointwise limit. Why this step? For $x \neq = 0$ , as $n \to \infty$ the $\frac{1}{n} \to 0$ , giving $\frac{x ^{2}}{x ^{2}} = 1$ . At $x = 0$ we always have $0$ . So $f (x) = {10 x \neq = 0 x = 0,$ a removable-looking discontinuity at $0$ .

Step 2. The degenerate point. Why this step? We must confirm the sup isn't accidentally tamed by $x = 0$ . At $x = 0$ , $∣ f_{n} - f ∣ = ∣0 - 0∣ = 0$ — harmless. The danger is near $0$ , not at it.

Step 3. Compute $M_{n}$ near the trouble point. Why this step? For small $x \neq = 0$ , $f (x) = 1$ but $f_{n} (x) = \frac{x ^{2}}{x ^{2} + 1/ n}$ can be made close to $0$ by taking $x$ tiny (e.g. $x = \frac{1}{n}$ gives $\frac{1/ n}{1/ n + 1/ n} = \frac{1}{2}$ , so the gap is $\frac{1}{2}$ ). Hence $M_{n} \geq \frac{1}{2}$ for all $n$ .

Step 4. Verdict. $M_{n} \neq \to 0$ : not uniform on $R$ (and the discontinuous limit already told us so via Cell B). ❌

Verify: The isolated degenerate point $x = 0$ contributes gap $0$ and is a red herring; the real killer is the neighbourhood around it. Sanity: at $x = 1/ n$ , $f_{n} = 1/2$ exactly, gap $= 1/2$ , confirmed for any $n$ .

Example 6 — Series via the M-test (Cell G)

Worked example

$S (x) = k = 1 \sum \infty \frac{cos ( k x )}{k ^{2}}$ on $R$ ; show the partial sums converge uniformly. Forecast: We do not know $S (x)$ in closed form. Can we still prove uniform convergence? Guess which tool sidesteps needing the limit.

Step 1. Bound each term independently of $x$ . Why this step? The Weierstrass M-test says: if $∣ u_{k} (x) ∣ \leq M_{k}$ for all $x$ and $\sum M_{k} < \infty$ , the series converges uniformly. We need such constants $M_{k}$ . Since $∣ cos (k x) ∣ \leq 1$ , $\frac{c o s ( k x )}{k ^{2}} \leq \frac{1}{k ^{2}} =: M_{k} .$

Step 2. Check the constant series converges. Why this step? The M-test needs $\sum M_{k} < \infty$ . Here $\sum_{k \geq 1} \frac{1}{k ^{2}} = \frac{π ^{2}}{6} < \infty$ (the Basel sum).

Step 3. Conclude. Why this step? Both hypotheses hold, so $\sum \frac{c o s k x}{k ^{2}}$ converges uniformly on all of $R$ . Because each term is continuous, so is $S$ . ✅

Verify: No pointwise limit was ever computed — that is the whole power of the M-test (it is really the uniform Cauchy criterion in disguise: the tail $\sum_{k > N} \frac{1}{k ^{2}}$ is a uniform, $x$ -free bound). Numerically the tail from $k = 11$ onward is $\leq \sum_{k = 11}^{\infty} k^{- 2} \approx 0.0952$ , and it keeps shrinking.

Example 7 — Real-world "error budget" framing (Cell H)

Worked example A signal reconstruction sends approximations

$f_{n} (t) = f (t) + \frac{r}{n} sin (ω t)$ to every receiver at time-index $t$ . Engineering spec: the error must be below $ε = 0.001$ for all receivers simultaneously. Take amplitude $r = 2$ . After which $n$ is the spec met? Forecast: Is there one deadline $N$ that satisfies all $t$ at once, or must each $t$ wait its own time?

Step 1. Identify the deviation. Why this step? "Error at $t$ " is $∣ f_{n} (t) - f (t) ∣ = \frac{r}{n} ∣ sin (ω t) ∣$ . The worst receiver is where $∣ sin ω t ∣ = 1$ , so $M_{n} = sup_{t} \frac{r}{n} ∣ sin ω t ∣ = \frac{r}{n} = \frac{2}{n} .$ This bound is free of $t$ — a uniform situation (Cell A structurally), which is exactly why a single deadline can exist.

Step 2. Solve the budget. Why this step? We need $M_{n} < ε$ : $\frac{2}{n} < 0.001 ⟹ n > 2000$ . So $n \geq 2001$ .

Step 3. Interpret. One shared $N = 2001$ satisfies every receiver — the "single stopwatch" picture from the parent's Feynman story. ✅

Verify: At $n = 2001$ , worst-case error $= \frac{2}{2001} \approx 9.995 \times 1 0^{- 4} < 1 0^{- 3}$ . At $n = 2000$ , $\frac{2}{2000} = 1 0^{- 3}$ , not strictly below, so $2000$ fails the strict spec — the boundary matters.

Example 8 — Exam twist: uniform yet derivatives explode (Cell I)

Worked example

$f_{n} (x) = \frac{sin ( n ^{2} x )}{n}$ on $R$ . Is $f_{n} \to f$ uniform? Does $f_{n}^{'} \to f^{'}$ ? Forecast: The amplitude $\frac{1}{n}$ shrinks — smells uniform. But the frequency $n^{2}$ grows fast. Guess what happens to the slopes.

Step 1. Uniform convergence of $f_{n}$ . Why this step? $∣ f_{n} (x) ∣ \leq \frac{1}{n}$ for all $x$ , so $f = 0$ pointwise and $M_{n} = sup_{x} \frac{∣ s i n n ^{2} x ∣}{n} = \frac{1}{n} \to 0$ . Uniform. ✅

Step 2. The derivative sequence. Why this step? Differentiation multiplies by the inner frequency: $f_{n}^{'} (x) = \frac{n ^{2}}{n} cos (n^{2} x) = n cos (n^{2} x)$ . At $x = 0$ : $f_{n}^{'} (0) = n \to \infty$ . So $f_{n}^{'}$ diverges, badly. ❌

Step 3. Moral. Why this step? Uniform convergence of $f_{n}$ says nothing about $f_{n}^{'}$ ; differentiation amplifies high-frequency wiggles (parent's third mistake). To conclude $f_{n}^{'} \to f^{'}$ you need a separate theorem: uniform convergence of $f_{n}^{'}$ plus convergence at one point.

Verify: $f_{n} \to 0$ uniformly (amplitude $1/ n$ ) but $f_{n}^{'} (0) = n \to \infty$ . Contrast with $\int$ : since convergence is uniform, integrals do pass to the limit — it is only differentiation that fails.

Recall

Recall Which cell, which test? (cover the answers)

Limit turns out discontinuous — verdict? ::: Automatically not uniform (Cell B), no sup computation needed.
Pointwise limit is $0$ but a bump travels — what is $M_{n}$ often? ::: A constant like $e^{- 1}$ that never $\to 0$ (Cell C/F); find the peak with a derivative.
Same formula fails on $E$ but holds on a shrunk $E^{'}$ — name it. ::: Domain-dependence (Cell D); exclude the trouble point.
Series, no closed-form limit — tool? ::: Weierstrass M-test / uniform Cauchy (Cell G).
Uniform $f_{n}$ but wild $f_{n}^{'}$ — allowed? ::: Yes (Cell I); differentiation needs its own theorem.

Connections

Parent: Uniform convergence
Pointwise convergence — always computed first.
Weierstrass M-test — Example 6.
Continuity preserved under uniform limits — Example 2's shortcut.
Interchange of limit and integral / Dominated convergence theorem — Example 8's contrast.
Cauchy sequences in metric spaces — the M-test's engine.
Equicontinuity and Arzelà–Ascoli.