Uniform convergence of function sequences — Hinglish notes

4.10.24 · Maths › Advanced Topics (Elite Level)

1. Definitions, built from scratch

2. The sup-norm criterion (the working tool)

Figure — Uniform convergence of function sequences

3. The classic counterexample (must memorize)

Worked example

$f_{n} (x) = x^{n}$ on $[0, 1]$ Step 1 — pointwise limit dhundo. Kyun? Uniformity test se pehle $f$ jaanna zaroori hai.

$0 \leq x < 1$ ke liye: $x^{n} \to 0$ .
$x = 1$ ke liye: $1^{n} = 1 \to 1$ . Toh $f (x) = {01 0 \leq x < 1 x = 1$ .

Step 2 — note karo ki $f$ discontinuous hai $x = 1$ par, jabki har $f_{n}$ continuous hai. Why this matters: agar convergence uniform hoti, toh limit continuous honi chahiye (Theorem §4 mein). Yeh break hamare liye clue hai ki yeh uniform NAHI hai.

Step 3 — $M_{n}$ compute karo. Kyun? Direct sup test. $M_{n} = sup_{x \in [0, 1]} ∣ x^{n} - f (x) ∣.$ $x \in [0, 1)$ ke liye gap $x^{n}$ hai, jiska sup jab $x \to 1^{-}$ jaata hai toh $1$ hota hai. Toh $M_{n} = 1$ sabhi $n$ ke liye. Kyunki $M_{n} = 1 \neq \to 0$ , convergence uniform nahi hai. $■$

Forecast-then-verify: Calculate karne se pehle, shayad tumhara guess ho "yeh $0$ pe nicely converge karta hai." Asli problem woh point hai jo $x = 1$ ki taraf sneakily jaata hai, jahaan $x^{n}$ $1$ ke paas rehta hai. Sup us point ko pakad leta hai.

4. Why uniform convergence is worth the trouble

5. Cauchy criterion (test without knowing the limit)

Common mistakes

Recall & Feynman

Recall Active recall — cover the answers

Dono definitions state karo; ek hi kya FARK hai? ::: Quantifier order; $N (ε, x)$ vs $N (ε)$ .
Uniformity ke liye single computational test? ::: $M_{n} = sup_{x} ∣ f_{n} - f ∣ \to 0$ .
$x^{n}$ on $[0, 1]$ uniform kyun nahi hai? ::: $M_{n} = 1$ hamesha; limit $1$ par discontinuous hai.

Recall Feynman: explain to a 12-year-old

Socho runners ki ek class (har $x$ ek runner hai) sab ek finish line $f (x)$ ki taraf ja rahe hain. Pointwise: har runner eventually aa jaata hai, lekin kuch super slow hain — tum nahi keh sakte "10 minutes mein sab done." Uniform: ek single stopwatch time hai jiske baad har runner already finish ke tiny circle ke andar hai. Ek time sab ke liye fit. Yahi shared deadline hai jo tumhe poori crowd ke baare mein safely nice baatein kehne deta hai.

Connections

Pointwise convergence — iska weaker cousin.
Weierstrass M-test — series ki uniform convergence ke liye sufficient condition.
Continuity preserved under uniform limits — $ε /3$ theorem.
Interchange of limit and integral / Dominated convergence theorem — measure-theoretic upgrade.
Cauchy sequences in metric spaces — uniform Cauchy = sup-norm ke under $C (E)$ ki completeness.
Equicontinuity and Arzelà–Ascoli.

Pointwise vs uniform — key difference kya hai

Quantifiers ka order: pointwise mein

N = N (ε, x)

allowed hai; uniform mein sabhi

x

ke liye ek

N = N (ε)

chahiye.

Uniform convergence ke liye sup-norm test

f_{n} \to f

uniformly

⟺ M_{n} = sup_{x} ∣ f_{n} (x) - f (x) ∣ \to 0

Kya

x^{n}

[0, 1]

par uniformly convergent hai?

Nahi;

M_{n} = 1

sabhi

n

ke liye aur limit

x = 1

par jump karti hai.

Kya

x^{n}

[0, a]

a < 1

par uniformly convergent hai?

Haan;

M_{n} = a^{n} \to 0

Kaun si property uniform (pointwise nahi) convergence preserve karti hai?

Limit function ki continuity (saath hi integral interchange bhi).

Counterexample ki uniform conv. of

f_{n}

conv. of

f_{n}^{'}

nahi deta

f_{n} = sin (n x) / n \to 0

uniformly lekin

f_{n}^{'} = cos (n x)

diverge karta hai.

Uniform Cauchy criterion ka statement

Uniform convergence

⟺ \forall ε \exists N : m, n \geq N \Rightarrow sup_{x} ∣ f_{n} - f_{m} ∣ < ε

Uniform conv. ke under

\int f_{n} \to \int f

kyun?

∣ \int f_{n} - \int f ∣ \leq (b - a) M_{n} \to 0