Uniform convergence of function sequences: Worked examples

4.10.24 · D3 · Maths › Advanced Topics (Elite Level) › Uniform convergence of function sequences

Scenario matrix

Har uniform-convergence exercise inhi cells mein se kisi ek mein aati hai. Neeche ke examples un cells ke saath tagged hain jo woh cover karte hain.

Cell	Situation	Kya galat ho sakta hai / kya dekhna hai
A	Limit continuous hai, convergence uniform hai	benign — $M_{n}$ ko ek $x$ se independent formula se bound karo
B	Limit discontinuous hai ⇒ kabhi uniform nahi	jump failure ki pehchaan hai
C	Ek travelling bump (max domain par chhup ke jaata hai)	$M_{n}$ constant reh sakta hai bhi jab $f \equiv 0$ ho
D	Uniformity domain par depend karti hai ( $E$ par fail, $E^{'} \subset E$ par hold)	trouble point se door ho jaao
E	Degenerate / zero input: $x = 0$ , endpoints, $f_{n}$ already $= f$	check karo ki trivial points sup ko todte toh nahi
F	Limiting behaviour: sup calculus se milta hai (derivative se max dhundho)	locate karo ki peak step $n$ par kahan hai
G	Series case — Cauchy / Weierstrass M-test use karo bina $f$ jaane	har term ko ek summable constant se bound karo
H	Word problem / real-world framing (ek "error budget" deadline)	"sab users ke liye ek saath" = uniform translate karo
I	Exam twist: convergence uniform hai lekin derivatives/integrals behave nahi karte	$f_{n}$ ka uniform ≠ $f_{n}^{'}$ ka uniform

Neeche ke chhe figures cells B, C, D, F ki geometry dikhate hain.

Example 1 — Benign uniform (Cell A)

Worked example

$f_{n} (x) = \frac{x}{n}$ on $[0, 1]$ Forecast: Abhi andaza lagao — kya poora graph ek shared pace se $0$ par flatten hota hai, ya koi point peechhe reh jaata hai?

Step 1. Pointwise limit nikalo. Yeh step kyun? Bina $f$ jaane hum gap measure nahi kar sakte. Har fixed $x$ ke liye, $\frac{x}{n} \to 0$ jab $n \to \infty$ , toh $f (x) = 0$ .

Step 2. Gap likho. Yeh step kyun? $M_{n}$ , $∣ f_{n} - f ∣$ ka sup hai. Yahan $∣ f_{n} (x) - f (x) ∣ = \frac{x}{n} - 0 = \frac{x}{n}$ .

Step 3. $x \in [0, 1]$ par sup lo. Yeh step kyun? $\frac{x}{n}$ , $x$ mein increase karta hai, toh sabse badi value right endpoint $x = 1$ par hai: $M_{n} = sup_{x \in [0, 1]} \frac{x}{n} = \frac{1}{n} .$

Step 4. $M_{n} \to 0$ test karo. $\frac{1}{n} \to 0$ , toh convergence uniform hai. ✅

Verify: Bound $\frac{1}{n}$ mein $x$ ka koi zikr nahi — yahi uniformity ki pehchaan hai. Ek value sanity check karo: $n = 100$ par, koi bhi point $0.01$ se zyada deviate nahi karta; poora graph half-width $0.01$ ke ek horizontal band ke andar hai.

Example 2 — Discontinuous limit uniformity ko khatam karta hai (Cell B)

Worked example

$f_{n} (x) = \frac{1}{1 + x ^{2 n}}$ on $[0, 2]$ Forecast: Har $f_{n}$ ek smooth bump hai. Limit ki shape guess karo — kya woh smooth rehti hai?

Step 1. Pointwise limit — $x$ ki size ke hisaab se split karo. Yeh step kyun? $x^{2 n}$ ka behavior $x < 1$ , $x = 1$ , $x > 1$ ke liye bilkul alag hai, toh hume case-split karni hogi.

$0 \leq x < 1$ : $x^{2 n} \to 0$ , toh $f_{n} \to \frac{1}{1 + 0} = 1$ .
$x = 1$ : $x^{2 n} = 1$ , toh $f_{n} = \frac{1}{2}$ sab $n$ ke liye; limit $= \frac{1}{2}$ .
$1 < x \leq 2$ : $x^{2 n} \to \infty$ , toh $f_{n} \to 0$ .

Toh $f (x) = ⎩ ⎨ ⎧ 1 \frac{1}{2} 0 0 \leq x < 1 x = 1 1 < x \leq 2$ — $x = 1$ par ek jump hai.

Step 2. Continuity theorem ka contrapositive lagao. Yeh step kyun? Har $f_{n}$ continuous hai. Agar convergence uniform hoti, toh limit bhi continuous hoti. Woh nahi hai — $x = 1$ par jump karta hai. Isliye convergence uniform nahi hai. ❌

Step 3. $M_{n}$ se directly confirm karo. Yeh step kyun? Hume ek doosra, computational witness chahiye. $x \to 1^{-}$ se approach karo: wahan $f (x) = 1$ hai lekin $f_{n} (x) = \frac{1}{1 + x ^{2 n}}$ , $\frac{1}{2}$ ki taraf dip karta hai. Exactly $x = 1$ par, $∣ f_{n} (1) - f (1) ∣ = ∣ \frac{1}{2} - \frac{1}{2} ∣ = 0$ , lekin $1$ se thoda left mein gap $\frac{1}{2}$ tak chadh jaata hai. Toh $M_{n} \geq \frac{1}{2}$ sab $n$ ke liye, isliye $M_{n} \neq \to 0$ .

Verify: Do independent arguments (discontinuity + sup) agree karte hain. Figure mein dekho: curves $1$ ke paas rehti hain phir plunge karti hain, aur woh plunge kabhi ek step mein sharpen nahi hoti jise tum ek patle band mein trap kar sako.

Example 3 — Travelling bump (Cell C, F)

Worked example

$f_{n} (x) = n x e^{- n x}$ on $[0, \infty)$ Forecast: Har fixed $x > 0$ ke liye yeh $\to 0$ jaata hai (exponential factor $n$ ko beat karta hai). Toh $f \equiv 0$ . Kya iska matlab uniform hai? Aage padhne se pehle guess karo.

Step 1. Pointwise limit. Yeh step kyun? $x = 0$ par, $f_{n} (0) = 0$ . Fixed $x > 0$ ke liye, $e^{- n x}$ exponentially decay karta hai jabki $n x$ sirf linearly badhta hai, toh $n x e^{- n x} \to 0$ . Isliye $f (x) = 0$ har jagah.

Step 2. Derivative se $f_{n}$ ki peak locate karo. Yeh step kyun? $M_{n} = sup_{x} n x e^{- n x}$ ; graph sabse upar kahan hai yeh jaanne ke liye derivative zero karte hain — yahi calculus tool hai jo answer deta hai "maximum kahan hai?" $\frac{d}{d x} (n x e^{- n x}) = n e^{- n x} (1 - n x) = 0 ⟹ x_{n} = \frac{1}{n} .$ Peak $x_{n} = \frac{1}{n}$ par hai, jo $n$ badhne ke saath $0$ ki taraf slide karta hai — ek moving bump.

Step 3. Peak ki height. Yeh step kyun? $x_{n} = \frac{1}{n}$ wapas plug karo: $M_{n} = n \cdot \frac{1}{n} \cdot e^{- n \cdot \frac{1}{n}} = e^{- 1} \approx 0.368.$

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

Verify: Figure dekho — har curve ki height same $1/ e$ hai, bas peak left ki taraf march kar rahi hai. $1/ e$ se kam height ka koi bhi horizontal band kabhi saari tails contain nahi kar sakta. Numerically $e^{- 1} = 0.3679$ , $n$ mein constant.

Example 4 — Uniformity domain par depend karti hai (Cell D)

Worked example Same

$f_{n} (x) = n x e^{- n x}$ , ab $[δ, \infty)$ par fixed $δ > 0$ ke saath Forecast: Hum abhi $[0, \infty)$ par fail ho gaye kyunki bump $x = 0$ tak pahunchti hai. Agar hum $x$ ko $0$ ke paas jaane se rok dein, toh kya hum jeet sakte hain?

Step 1. Peak $δ$ ke relative kahan hai? Yeh step kyun? Unconstrained peak $x_{n} = \frac{1}{n}$ par hai. Jab $n > \frac{1}{δ}$ (yaani $\frac{1}{n} < δ$ ), peak domain se bahar slide ho jaati hai — woh $δ$ ke left mein hai. $[δ, \infty)$ par function $n x e^{- n x}$ tab decreasing hai, toh uska sup left endpoint $x = δ$ par hai.

Step 2. Sup evaluate karo. Yeh step kyun? $n > \frac{1}{δ}$ ke liye, $M_{n} = n δ e^{- n δ} .$

Step 3. Limit lo. Yeh step kyun? $e^{- n δ}$ , $n$ mein exponentially decay karta hai aur linear $n δ$ ko crush kar deta hai, toh $n δ e^{- n δ} \to 0$ . $[δ, \infty)$ par Uniform. ✅

Verify: Same formula, ulta verdict — yahi moral hai parent ke $x^{n}$ examples se: uniformity (function, domain) pair ki ek property hai. Numerically $δ = 1$ ke saath: $M_{5} = 5 e^{- 5} \approx 0.0337$ , $M_{10} = 10 e^{- 10} \approx 4.54 \times 1 0^{- 4}$ — clearly $0$ ki taraf ja raha hai.

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

Worked example

$f_{n} (x) = \frac{x ^{2}}{x ^{2} + \frac{1}{n}}$ on $R$ , $x = 0$ par special dhyan ke saath Forecast: $x = 0$ par formula $\frac{0}{0 + 1/ n} = 0$ deta hai har $n$ ke liye — ek point jahan $f_{n}$ kabhi hilta nahi. Kya woh akela point trouble deta hai?

Step 1. Pointwise limit. Yeh step kyun? $x \neq = 0$ ke liye, $n \to \infty$ mein $\frac{1}{n} \to 0$ , toh $\frac{x ^{2}}{x ^{2}} = 1$ milta hai. $x = 0$ par hamesha $0$ hai. Toh $f (x) = {10 x \neq = 0 x = 0,$ $0$ par ek removable-looking discontinuity hai.

Step 2. Degenerate point. Yeh step kyun? Hume confirm karna hai ki sup accidentally $x = 0$ se tame toh nahi ho gaya. $x = 0$ par, $∣ f_{n} - f ∣ = ∣0 - 0∣ = 0$ — harmless. Danger $0$ ke paas hai, par nahi.

Step 3. Trouble point ke paas $M_{n}$ compute karo. Yeh step kyun? Chhote $x \neq = 0$ ke liye, $f (x) = 1$ hai lekin $f_{n} (x) = \frac{x ^{2}}{x ^{2} + 1/ n}$ ko $x$ tiny leke $0$ ke close kiya ja sakta hai (e.g. $x = \frac{1}{n}$ deta hai $\frac{1/ n}{1/ n + 1/ n} = \frac{1}{2}$ , toh gap $\frac{1}{2}$ hai). Isliye $M_{n} \geq \frac{1}{2}$ sab $n$ ke liye.

Step 4. Verdict. $M_{n} \neq \to 0$ : $R$ par not uniform (aur discontinuous limit ne Cell B ke through pehle hi bata diya tha). ❌

Verify: Isolated degenerate point $x = 0$ gap $0$ contribute karta hai aur red herring hai; asli killer uske aas-paas ka neighbourhood hai. Sanity: $x = 1/ n$ par, $f_{n} = 1/2$ exactly, gap $= 1/2$ , kisi bhi $n$ ke liye confirmed.

Example 6 — M-test se Series (Cell G)

Worked example

$S (x) = k = 1 \sum \infty \frac{cos ( k x )}{k ^{2}}$ on $R$ ; dikhao ki partial sums uniformly converge karte hain. Forecast: Hume $S (x)$ closed form mein pata nahi. Kya hum phir bhi uniform convergence prove kar sakte hain? Guess karo kaunsa tool limit jaane ki zaroorat ko bypass karta hai.

Step 1. Har term ko $x$ se independent bound karo. Yeh step kyun? Weierstrass M-test kehta hai: agar $∣ u_{k} (x) ∣ \leq M_{k}$ sab $x$ ke liye aur $\sum M_{k} < \infty$ , toh series uniformly converge karti hai. Hume aisi constants $M_{k}$ chahiye. Kyunki $∣ cos (k x) ∣ \leq 1$ , $\frac{c o s ( k x )}{k ^{2}} \leq \frac{1}{k ^{2}} =: M_{k} .$

Step 2. Check karo ki constant series converge kare. Yeh step kyun? M-test ko $\sum M_{k} < \infty$ chahiye. Yahan $\sum_{k \geq 1} \frac{1}{k ^{2}} = \frac{π ^{2}}{6} < \infty$ (Basel sum).

Step 3. Conclude karo. Yeh step kyun? Dono hypotheses hold karti hain, toh $\sum \frac{c o s k x}{k ^{2}}$ poore $R$ par uniformly converge karta hai. Kyunki har term continuous hai, $S$ bhi continuous hai. ✅

Verify: Koi bhi pointwise limit kabhi compute nahi ki — yahi M-test ki poori power hai (yeh actually uniform Cauchy criterion disguise mein hai: tail $\sum_{k > N} \frac{1}{k ^{2}}$ ek uniform, $x$ -free bound hai). Numerically $k = 11$ se aage ka tail $\leq \sum_{k = 11}^{\infty} k^{- 2} \approx 0.0952$ hai, aur yeh shrink karta rehta hai.

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

Worked example Ek signal reconstruction approximations

$f_{n} (t) = f (t) + \frac{r}{n} sin (ω t)$ har receiver ko time-index $t$ par bhejta hai. Engineering spec: error $ε = 0.001$ se kam honi chahiye sab receivers ke liye simultaneously. Amplitude $r = 2$ lo. Kis $n$ ke baad spec meet hoti hai? Forecast: Kya ek deadline $N$ hai jo sab $t$ ke liye satisfy ho, ya har $t$ ko apna alag time wait karna padta hai?

Step 1. Deviation identify karo. Yeh step kyun? " $t$ par error" hai $∣ f_{n} (t) - f (t) ∣ = \frac{r}{n} ∣ sin (ω t) ∣$ . Worst receiver woh hai jahan $∣ sin ω t ∣ = 1$ , toh $M_{n} = sup_{t} \frac{r}{n} ∣ sin ω t ∣ = \frac{r}{n} = \frac{2}{n} .$ Yeh bound $t$ se free hai — ek uniform situation (structurally Cell A), isliye ek single deadline exist kar sakti hai.

Step 2. Budget solve karo. Yeh step kyun? Hume $M_{n} < ε$ chahiye: $\frac{2}{n} < 0.001 ⟹ n > 2000$ . Toh $n \geq 2001$ .

Step 3. Interpret karo. Ek shared $N = 2001$ , har receiver ko satisfy karta hai — parent ki Feynman story se "single stopwatch" ki picture. ✅

Verify: $n = 2001$ par, worst-case error $= \frac{2}{2001} \approx 9.995 \times 1 0^{- 4} < 1 0^{- 3}$ . $n = 2000$ par, $\frac{2}{2000} = 1 0^{- 3}$ , strictly kam nahi hai, toh $2000$ strict spec fail karta hai — boundary matter karti hai.

Example 8 — Exam twist: uniform hai lekin derivatives explode karte hain (Cell I)

Worked example

$f_{n} (x) = \frac{sin ( n ^{2} x )}{n}$ on $R$ . Kya $f_{n} \to f$ uniform hai? Kya $f_{n}^{'} \to f^{'}$ ? Forecast: Amplitude $\frac{1}{n}$ shrink hota hai — uniform lagta hai. Lekin frequency $n^{2}$ fast badhti hai. Slopes ka kya hoga guess karo.

Step 1. $f_{n}$ ki uniform convergence. Yeh step kyun? $∣ f_{n} (x) ∣ \leq \frac{1}{n}$ sab $x$ ke liye, toh $f = 0$ pointwise aur $M_{n} = sup_{x} \frac{∣ s i n n ^{2} x ∣}{n} = \frac{1}{n} \to 0$ . Uniform. ✅

Step 2. Derivative sequence. Yeh step kyun? Differentiation inner frequency se multiply karta hai: $f_{n}^{'} (x) = \frac{n ^{2}}{n} cos (n^{2} x) = n cos (n^{2} x)$ . $x = 0$ par: $f_{n}^{'} (0) = n \to \infty$ . Toh $f_{n}^{'}$ diverge karta hai, buri tarah se. ❌

Step 3. Moral. Yeh step kyun? $f_{n}$ ki uniform convergence $f_{n}^{'}$ ke baare mein kuch nahi kehti; differentiation high-frequency wiggles ko amplify karta hai (parent ki teesri galti). $f_{n}^{'} \to f^{'}$ conclude karne ke liye ek alag theorem chahiye: $f_{n}^{'}$ ki uniform convergence plus ek point par convergence.

Verify: $f_{n} \to 0$ uniformly (amplitude $1/ n$ ) lekin $f_{n}^{'} (0) = n \to \infty$ . $\int$ se contrast karo: kyunki convergence uniform hai, integrals limit ke paas jaate hain — sirf differentiation fail karta hai.

Recall

Recall Kaunsi cell, kaunsa test? (answers cover karo)

Limit discontinuous nikle — verdict? ::: Automatically not uniform (Cell B), koi sup computation zaroor nahi.
Pointwise limit $0$ hai lekin bump travel karta hai — $M_{n}$ aksar kya hota hai? ::: Ek constant jaise $e^{- 1}$ jo kabhi $\to 0$ nahi karta (Cell C/F); derivative se peak dhundho.
Same formula $E$ par fail, chhote $E^{'}$ par hold — iska naam kya hai? ::: Domain-dependence (Cell D); trouble point ko exclude karo.
Series, closed-form limit nahi — tool kya hai? ::: Weierstrass M-test / uniform Cauchy (Cell G).
Uniform $f_{n}$ lekin wild $f_{n}^{'}$ — allowed hai? ::: Haan (Cell I); differentiation ko apna alag theorem chahiye.

Connections

Parent: Uniform convergence
Pointwise convergence — hamesha pehle compute hota hai.
Weierstrass M-test — Example 6.
Continuity preserved under uniform limits — Example 2 ka shortcut.
Interchange of limit and integral / Dominated convergence theorem — Example 8 ka contrast.
Cauchy sequences in metric spaces — M-test ka engine.
Equicontinuity and Arzelà–Ascoli.