Uniform convergence of function sequences: Question bank

4.10.24 · D5 · Maths › Advanced Topics (Elite Level) › Uniform convergence of function sequences

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Quantifier order. Pointwise hai $\forall x \forall ε \exists N$ ; uniform hai $\forall ε \exists N \forall x$ . Poore graph ke liye ek hi $N$ .
Sup test. $M_{n} = sup_{x \in E} ∣ f_{n} (x) - f (x) ∣$ ; uniform $⟺ M_{n} \to 0$ .

True or false — justify

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Agar $f_{n} \to f$ uniformly on $E$ , toh $f_{n} \to f$ pointwise on $E$ . ::: True. Uniform ek aisa $N$ deta hai jo sab $x$ ke liye kaam karta hai, toh specifically har fixed $x$ pe bhi kaam karta hai — yahi exactly pointwise hai. Uniform strictly stronger hai, kabhi weaker nahi. Agar $f_{n} \to f$ pointwise aur har $f_{n}$ continuous hai, toh $f$ continuous hai. ::: False. $x^{n}$ on $[0, 1]$ pointwise aise function pe converge karta hai jo $x = 1$ pe jump karta hai. Tumhe uniform convergence chahiye $ε /3$ argument ke liye. $f_{n} (x) = x^{n}$ har fixed $a < 1$ ke liye $[0, a]$ pe uniformly converge karta hai. ::: True. Wahan sup right endpoint pe attain hota hai, $M_{n} = a^{n} \to 0$ . Trouble point $x = 1$ exclude hai, isliye uniformity restore ho jaati hai. Har $a < 1$ ke liye $[0, a]$ pe uniform convergence, $[0, 1)$ pe uniform convergence imply karta hai. ::: False. "Har piece pe uniform" ka matlab "union pe uniform" nahi hai. Jaise $a \to 1$ , required $N$ blow up karta hai; poore $[0, 1)$ pe $x^{n}$ ka sup har $n$ ke liye $1$ hai. Agar $M_{n} \to 0$ toh $f_{n} \to f$ uniformly, chahe kaun sa bhi $f$ use karo. ::: False as stated — tumhe sahi pointwise limit $f$ use karni chahiye. Koi aur target function use karne se $M_{n}$ galat quantity ban jaati hai, aur ho sakta hai wo $0$ pe bhi na jaaye. $f_{n}$ ki uniform convergence $f_{n}^{'}$ ki uniform convergence imply karta hai. ::: False. $\frac{s i n ( n x )}{n} \to 0$ uniformly, lekin $f_{n}^{'} = cos (n x)$ bilkul converge nahi karta. Differentiation fast wiggles ko amplify karta hai. Agar $f_{n} \to f$ aur $g_{n} \to g$ dono $E$ pe uniformly hain, toh $f_{n} + g_{n} \to f + g$ uniformly. ::: True. $sup ∣ (f_{n} + g_{n}) - (f + g) ∣ \leq sup ∣ f_{n} - f ∣ + sup ∣ g_{n} - g ∣ \to 0$ triangle inequality on the sup se. Agar $f_{n} \to f$ aur $g_{n} \to g$ $E$ pe uniformly hain, toh $f_{n} g_{n} \to f g$ $E$ pe uniformly. ::: Generally false. Products ke liye boundedness chahiye: unbounded $E$ pe, $f_{n} (x) = x + \frac{1}{n}$ times $g_{n} = x$ dikhata hai ki error $\frac{x}{n}$ ka infinite sup hai. Ye true hai jab limits bounded hoon. Uniform Cauchy criterion se tum limit jaane bina uniform convergence prove kar sakte ho. ::: True. Agar $sup_{x} ∣ f_{n} - f_{m} ∣ \to 0$ as $m, n \to \infty$ , toh sequence kisi $f$ pe uniformly converge karti hai; sup-norm space ki completeness limit supply karti hai. Uniform convergence functions ki boundedness preserve karti hai. ::: True (large $n$ ke liye). Agar $∣ f_{n} - f ∣ < 1$ sab $x$ ke liye jab $n \geq N$ aur $f$ bounded hai, toh har aisa $f_{n}$ bounded hai; aur $f$ khud bhi bounded hota hai jab $f_{n}$ uniformly bounded hoon.

Spot the error

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" $f_{n} (x) = x^{n}$ $[0, 1]$ pe $0$ pe converge karta hai, isliye $\int_{0}^{1} f_{n} \to \int_{0}^{1} 0 = 0$ ." ::: Integral conclusion numerically sahi hai ( $\int_{0}^{1} x^{n} = \frac{1}{n + 1} \to 0$ ), lekin justification galat hai: convergence uniform nahi hai, isliye tum interchange theorem invoke nahi kar sakte. Value kisi aur reason se agree karta hai. "Convergence uniform hai kyunki har $x$ ke liye maine ek $N (x)$ find kar liya." ::: Har point ke liye $N$ find karna sirf pointwise hai. Uniform ko ek single $N$ chahiye jo $x$ se independent ho — tumhe dikhana hoga ki $sup_{x} N (x)$ finite hai, yaani error ko uniformly bound karo. " $M_{n} = sup_{x \in [0, 1)} x^{n} = x^{n}$ largest $x$ pe evaluate hota hai, lekin $[0, 1)$ ka koi largest point nahi, isliye $M_{n}$ undefined hai." ::: Sup ka attain hona zaruri nahi. $sup_{x \in [0, 1)} x^{n} = 1$ as a supremum (jaise $x \to 1^{-}$ approach hota hai) chahe koi $x$ use hit na kare. Sup ≠ max. "Kyunki $∣ f_{n} (x) - f (x) ∣ < ε$ sab $x$ ke liye, hamare paas $M_{n} < ε$ hai, aur equality kabhi nahi ho sakti." ::: Tum sirf $M_{n} \leq ε$ conclude kar sakte ho; strict inequality sup lene par $\leq$ mein degrade ho sakti hai. Isliye definition-to-sup proof $ε /2$ use karta hai, taaki $M_{n} \leq ε /2 < ε$ . "Continuity ka $ε /3$ proof sirf $f_{n}$ ki continuity use karta hai, uniformity nahi." ::: Isne secretly step 1 mein uniformity use ki: same $N$ ko $∣ f - f_{N} ∣$ ko dono $x$ aur $x_{0}$ par ek saath control karna hai. Pointwise do alag $N$ deta aur argument collapse ho jaata. " $f_{n} \to f$ pointwise with $f$ continuous, isliye convergence uniform hai." ::: Continuous limit necessary hai lekin sufficient nahi. $f_{n} (x) = \frac{n x}{1 + n ^{2} x ^{2}}$ on $[0, 1]$ pointwise continuous function $0$ pe converge karta hai, lekin peak height $\frac{1}{2}$ pe rehta hai, isliye $M_{n} \neq \to 0$ . "Uniform convergence on $R$ ka matlab hai graphs eventually coincide kar jaate hain." ::: Wo ek shrinking horizontal band of width $2 ε$ ke andar rehte hain, identical nahi. "Coincide" ka matlab hota $f_{n} = f$ exactly; uniform sirf kehta hai maximum vertical gap $0$ pe shrink hota hai.

Why questions

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Quantifiers ko $\forall ε \exists N \forall x$ mein swap karna uniform ko pointwise se strictly harder kyun banata hai? ::: $\exists N$ ko $\forall x$ se pehle rakhna ek single $N$ ko simultaneously har point serve karne par majboor karta hai, isliye use worst point bhi handle karna padta hai. Pointwise slow points ko apna bada $N$ choose karne deta hai. " $sup$ of the gap $\to 0$ " natural test kyun hai, har point check karne ki jagah? ::: Phrase "sab $x$ ke liye, gap $< ε$ " is statement hai ki sabse bada gap $< ε$ hai, aur sabse bada gap precisely supremum hai. Isliye uniform convergence infinitely many point conditions ko ek number $M_{n}$ mein collapse karta hai. Uniform convergence limit aur integral interchange karne deta hai, lekin pointwise generally nahi — kyun? ::: Single bound $M_{n}$ poore integrand ko control karta hai, deta hai $∣ \int f_{n} - \int f ∣ \leq (b - a) M_{n} \to 0$ . Pointwise koi uniform bound nahi deta, isliye mass escape kar sakta hai (fixed area ka spike slide off ho sakta hai). Counterexample $sin (n x) / n$ kyun dikhata hai ki derivatives protected nahi hain? ::: Functions $\frac{1}{n}$ factor ki wajah se shrink karte hain, lekin differentiate karne se wo cancel ho jaata hai: $cos (n x)$ ka amplitude hamesha $1$ rehta hai. Heights ki uniform closeness slopes ke baare mein kuch nahi kehti. Uniform Cauchy convergence Weierstrass M-test jaise series ke liye useful kyun hai? ::: Tum $sup_{x} ∣ f_{n} - f_{m} ∣$ ko limit function identify karne se pehle ek numeric series ke tail sums se bound kar sakte ho, phir completeness limit deliver karti hai. Yahi exactly M-test uniform convergence certify karne ka tarika hai. Uniformity domain pe depend kyun karti hai, sirf formula pe nahi? ::: Sup domain ke upar liya jaata hai; domain enlarge karne se ek troublesome point expose ho sakta hai (jaise $x \to 1$ for $x^{n}$ ) jo $M_{n}$ inflate kar deta hai. Same formula, different sup, different verdict. Uniformity test karne se pehle pointwise limit pehle compute kyun karni padti hai? ::: $M_{n}$ define hota hai $sup ∣ f_{n} - f ∣$ ke roop mein, isliye tumhe literally $f$ chahiye use likhne ke liye. Galat $f$ guess karna galat gap measure karta hai aur meaningless answer deta hai.

Edge cases

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Kya empty set ya single point pe uniform convergence ka matlab hota hai? ::: Haan, trivially. Ek-point set pe pointwise aur uniform coincide karte hain (ek point khud se disagree nahi kar sakta), aur empty set pe empty set ke upar sup $0$ hai, isliye har sequence uniformly converge karti hai. Exactly boundary point $x = 1$ pe $f_{n} (x) = x^{n}$ ka kya hota hai? ::: $1^{n} = 1$ sab $n$ ke liye, isliye sequence constant $1$ hai wahan aur $1$ pe converge karti hai, $0$ pe nahi. Ye akela point hi pointwise limit ko discontinuous banata hai aur $[0, 1]$ pe uniformity destroy karta hai. Kya ek constant sequence $f_{n} = g$ (sab $n$ ke liye same function) uniformly convergent hai? ::: Haan. $M_{n} = sup ∣ g - g ∣ = 0$ har $n$ ke liye, isliye wo $g$ pe trivially uniformly converge karta hai — sabse strongest possible case. Agar har $f_{n}$ discontinuous hai lekin $f_{n} \to f$ uniformly, toh kya $f$ discontinuous honi chahiye? ::: Nahi. Continuity theorem one-directional hai: continuous $f_{n}$ continuous $f$ force karte hain, lekin discontinuous $f_{n}$ ka continuous (ya discontinuous) uniform limit ho sakta hai; uniformity koi break force nahi karta. Unbounded domain pe, kya ek sequence uniform ho sakti hai chahe har $f_{n}$ unbounded ho? ::: Haan. $f_{n} (x) = x + \frac{1}{n}$ on $R$ mein har $f_{n}$ unbounded hai, phir bhi $M_{n} = sup_{x} ∣ f_{n} - x ∣ = \frac{1}{n} \to 0$ , isliye wo $f (x) = x$ pe uniformly converge karta hai. $M_{n}$ kya hota hai jab pointwise limit kisi point pe exist nahi karta? ::: Toh $f$ wahan defined nahi hai, isliye definition stage pe hi uniform convergence fail ho jaati hai — koi target nahi jiske uniformly close ho sakein. Uniform convergence presuppose karta hai ki pointwise convergence har jagah ho. Degenerate case: agar $M_{n} \to 0$ lekin kabhi exactly $0$ nahi, kya wo phir bhi uniform hai? ::: Haan. Uniform convergence sirf $M_{n}$ ki limit $0$ hone ki demand karta hai; kabhi nahi kehta functions actually $f$ ke equal hon. $0$ pe tend karne wali strictly positive sequence perfectly uniform hai.

Connections

Pointwise convergence — wo weaker cousin jisse ye traps baar baar contrast karte hain.
Continuity preserved under uniform limits — $ε /3$ theorem jo kai items ke peeche hai.
Interchange of limit and integral / Dominated convergence theorem — jab limits swap karna legal ho.
Weierstrass M-test / Cauchy sequences in metric spaces — limit-free tests.
Equicontinuity and Arzelà–Ascoli — agla level upar.