Uniform continuity — difference from pointwise: Question bank

4.10.23 · D5 · Maths › Advanced Topics (Elite Level) › Uniform continuity — difference from pointwise

Drills se pehle, do players ka ek-line refresher jo tum compare kar rahe ho (poora banaya gaya hai parent note mein):

Recall Ek structural difference (agar zaroorat ho toh kholo)

Pointwise: $\forall x_{0} \forall ε \exists δ \forall x$ — $\exists δ$ point $x_{0}$ ke baad aata hai, isliye $δ = δ (ε, x_{0})$ us point ko dekh sakta hai. Uniform: $\forall ε \exists δ \forall x, y$ — $\exists δ$ points se pehle aata hai, isliye ek $δ = δ (ε)$ ko sabko serve karna padta hai. Quantifier order hi poori kahani hai.

True or false — justify

Inka rule yeh hai: sirf "true"/"false" likhne se kuch marks nahi milenge. Reason hi jawab hai.

x^{2}

har bounded interval

[0, M]

par uniformly continuous hai.

True —

[0, M]

par amplifier satisfy karta hai

∣ x + y ∣ \leq 2 M

, isliye

δ = ε / (2 M)

sab pairs ke liye kaam karta hai; slope-factor ki boundedness hi isko bachati hai.

x^{2}

[0, \infty)

par uniformly continuous hai.

False —

∣ x^{2} - y^{2} ∣ = ∣ x + y ∣∣ x - y ∣

mein factor

∣ x + y ∣

unbounded hai, isliye required gap

ε /∣ x + y ∣ \to 0

ho jaata hai aur koi ek

δ > 0

survive nahi karta.

Agar

f

R

par continuous hai, toh woh

R

par uniformly continuous hai.

False — continuity sirf har point par ek local

δ

ka waada karti hai; ek unbounded domain par un

δ

's ka infimum

0

ho sakta hai. Counterexample:

x^{2}

R

Agar

f

A

par uniformly continuous hai, toh woh

A

par (pointwise) continuous hai.

True — uniform continuity strictly stronger hai; jo ek

δ

sab pairs ke liye kaam karta hai woh specifically har fixed point par bhi kaam karta hai, isliye yeh pointwise continuity imply karta hai.

Har bounded domain continuity ko uniform continuity mein upgrade karne par majboor karta hai.

False — sirf bounded kaafi nahi; tumhe bounded aur closed (compact) chahiye. Bounded-but-open

(0, 1)

par,

1/ x

continuous hai phir bhi uniformly continuous nahi.

Ek uniformly continuous function ka derivative zaroor bounded hona chahiye.

False —

x

[0, \infty)

f^{'} (x) = 1/ (2 x) \to \infty

near

0

hai phir bhi

∣ x - y ∣ \leq ∣ x - y ∣

ke through uniformly continuous hai.

Lipschitz continuity, uniform continuity imply karta hai.

True —

∣ f (x) - f (y) ∣ \leq L ∣ x - y ∣

se,

δ = ε / L

choose karo; is

δ

mein koi

x

nahi hai, jo precisely uniformity hai.

Uniform continuity, Lipschitz continuity imply karta hai.

False —

x

uniformly continuous hai lekin Lipschitz nahi;

0

ke paas iska slope unbounded hai isliye koi ek

L

saare difference-quotients ko bound nahi kar sakta.

Compact set

[a, b]

par ek continuous function wahan hamesha uniformly continuous hota hai.

True — yeh Heine–Cantor hai:

[a, b]

ko local

δ

-balls se cover karo, finite subcover nikalo, minimum

δ > 0

lo.

A

par do uniformly continuous functions ka sum

A

par uniformly continuous hota hai.

True — agar

∣ x - y ∣ < δ

ke liye

∣ f (x) - f (y) ∣ < ε /2

aur

∣ g (x) - g (y) ∣ < ε /2

ho (do

δ

's mein se chhota use karo), toh triangle inequality deta hai

∣ (f + g) (x) - (f + g) (y) ∣ < ε

Ek unbounded domain par do uniformly continuous functions ka product uniformly continuous hota hai.

False —

f (x) = g (x) = x

dono

R

par (Lipschitz, hence) uniformly continuous hain, lekin

f g = x^{2}

R

par uniformly continuous nahi hai; unbounded functions ek doosre ko amplify kar sakte hain.

Ek uniformly continuous function har Cauchy sequence ko Cauchy sequence mein bhejta hai.

True — yeh ek defining strength hai:

ε

diya gaya, uska ek

δ

"terms eventually within

δ

" ko "images eventually within

ε

" mein badal deta hai. Sirf pointwise-continuous

f

isme fail ho sakta hai (e.g.

1/ x

(0, 1)

sin x

R

par uniformly continuous hai.

True —

∣ sin^{'} ∣ = ∣ cos ∣ \leq 1

, isliye yeh Lipschitz hai

L = 1

ke saath aur

δ = ε

sab jagah kaam karta hai chahe domain unbounded ho.

Spot the error

Har item ek plausible-lagta argument hai jisme ek hidden flaw hai. Woh flaw bolo.

(0, 1)

bounded hai, isliye iske upar continuity automatically uniform continuity deta hai."

Error yeh hai ki closed drop kar diya: Heine–Cantor ko closed aur bounded (compact) chahiye.

(0, 1)

open hai, aur iske upar

1/ x

continuous hai lekin uniformly continuous nahi.

x

ke liye

f^{'}

unbounded hai

[0, \infty)

par, isliye

x

uniformly continuous nahi hai."

Error yeh hai ki bounded-derivative ko necessary maan liya. Yeh sirf sufficient hai (MVE ke through).

x

ek genuine uniformly-continuous function hai jiska derivative unbounded hai.

x^{2}

ki uniform continuity disprove karne ke liye, maine

δ

aur

x

fix kiya aur dikhaya ki

∣ x^{2} - y^{2} ∣ > ε

y

ke liye

δ

ke andar; ho gaya."

Structure mein koi error nahi — lekin students jo galti karte hain woh hai

δ

ko last fix karna: tumhe adversary ko pehle koi bhi

δ

dena hoga, phir

x, y

produce karo. Parent ka choice

x = 1/ δ, y = 1/ δ + δ /2

exactly yahi karta hai.

f (x) = x

aur

g (x) = x

dono

R

par uniformly continuous hain, isliye

f \cdot g

bhi hai."

Error yeh hai ki yeh assume kar liya ki uniformly-continuous functions ke products uniformly continuous rehte hain. Unbounded domains par generally nahi rehte:

x \cdot x = x^{2}

fail karta hai.

"Maine

x^{2}

ke liye

δ = ε /∣ x + y ∣

choose kiya, jo sirf points par depend karta hai, isliye

x^{2}

uniformly continuous hai."

Error: ek valid uniform

δ

sirf

ε

par hi depend karna chahiye.

x, y

containing

δ

exactly pointwise (non-uniform) situation hai; yahan toh yeh

\to 0

bhi jaata hai jab

∣ x + y ∣ \to \infty

1/ x

(0, 1)

par Lipschitz hai kyunki wahan differentiable hai."

Differentiable

\neq =

bounded derivative. Yahan

f^{'} (x) = - 1/ x^{2} \to - \infty

x \to 0^{+}

, isliye koi ek

L

slopes ko bound nahi kar sakta;

1/ x

(0, 1)

par na Lipschitz hai na uniformly continuous.

"Infinitely many local

δ

's ka minimum ek positive uniform

δ

deta hai."

Error yeh hai ki infinite family par min liya — infimum

0

ho sakta hai. Compactness hi ise ek finite subcover tak reduce karti hai, jahan minimum genuinely positive hota hai.

Why questions

Kyun

\exists δ

aur

\forall x_{0}

ki swap donoN notions ke beech poora difference hai?

Kyunki yeh change karta hai ki

δ

kya dekh sakta hai: point ke baad (

δ (ε, x_{0})

) woh locally adapt kar sakta hai; points se pehle (

δ (ε)

) woh frozen hai aur domain ke worst pair ko survive karna padta hai.

Kyun

∣ x^{2} - y^{2} ∣ = ∣ x + y ∣∣ x - y ∣

factor karna

R

par failure turant reveal kar deta hai?

Yeh input gap

∣ x - y ∣

ko amplifier

∣ x + y ∣

se alag karta hai; jab amplifier unbounded hota hai, koi fixed input gap output ko har jagah

ε

se neeche nahi rakh sakta.

Kyun compactness (sirf boundedness nahi) continuity ko uniform continuity mein bachata hai?

Compactness local

δ

-balls ki ek infinite open cover ko finite subcover tak reduce karne deta hai, isliye minimum

δ

ek positive number hota hai (ek Lebesgue number); sirf boundedness aisi koi finite reduction nahi deti.

Kyun Lipschitz sufficient hai lekin uniform continuity ke liye necessary nahi?

Lipschitz clean

δ = ε / L

deta hai, lekin uniform continuity ko sirf koi bhi modulus chahiye jo gaps ko

ε

se jode;

x

ka modulus

∣ x - y ∣

kaam karta hai chahe koi linear bound

L ∣ x - y ∣

na ho.

Kyun uniform continuity Cauchy sequences preserve karta hai jabki pointwise continuity nahi karta?

Ek Cauchy sequence ke terms eventually kisi bhi

δ

ke andar aa jaate hain; ek single global

δ

phir images ko

ε

ke andar force karta hai. Global

δ

ke bina (sirf ek bure boundary point ke paas local wale), images spread ho sakte hain — e.g.

1/ x

on Cauchy sequence

1/ n

(0, 1)

Kyun ek function

[0, \infty)

jaisi unbounded domain par uniformly continuous ho sakta hai?

Kyunki uniformity outputs kitni tezi se separate ho sakte hain se concerned hai, domain size se nahi; agar increments controlled hain (e.g.

sin x

ke liye bounded slope, ya

x

ke liye concave flattening), toh ek

δ

kaafi hai chahe kitna bhi door jaao.

Edge cases

Kya ek constant function

f (x) = c

kisi bhi set par uniformly continuous hai?

Haan, degenerately —

∣ f (x) - f (y) ∣ = 0 < ε

har pair ke liye, isliye koi bhi

δ

kaam karta hai; yeh

L = 0

ke saath Lipschitz hai.

Kya

f (x) = 1/ x

[1, \infty)

par uniformly continuous hai?

Haan —

[1, \infty)

par

∣ f^{'} (x) ∣ = 1/ x^{2} \leq 1

hai, isliye yeh

L = 1

ke saath Lipschitz hai;

0

par wali takleef

1

se start karke exclude ho gayi hai.

Kya

x

tab bhi uniformly continuous rahta hai agar hum endpoint

0

include karein?

Haan —

∣ x - y ∣ \leq ∣ x - y ∣

δ = ε^{2}

saare

x, y \geq 0

ke liye kaam karta hai, jinmein

0

wale pairs bhi shaamil hain; endpoint koi issue nahi deta.

Kya

(a, b)

par uniform continuity

f

ko closed

[a, b]

par continuously extend karne deti hai?

Haan — uniform continuity

a

aur

b

ki taraf jaati Cauchy sequences ko Cauchy (hence convergent) image sequences mein map karta hai, endpoints par well-defined limit values deta hai.

"Local

δ

's ka min" ka kya hota hai jab ek interval non-compactness ki taraf jaata hai, maano

x^{2}

ke liye

[0, M]

with

M \to \infty

Kaam karne wala

δ = ε / (2 M)

0

tak shrink ho jaata hai; limit

[0, \infty)

mein koi positive

δ

nahi bachta — compactness kho jaane ki ek concrete tasveer.

Kya

x^{2}

, unbounded lekin "patley" set

{0} \cup [10, 10.5]

par uniformly continuous hai?

Haan — yeh set closed aur bounded (compact) hai, isliye Heine–Cantor seedha apply hota hai chahe gap kuch bhi ho; amplifier

∣ x + y ∣

wahan

21

se bound hai.

Connections

Continuity (pointwise) — woh weaker notion jiske against yahan har trap contrast karta hai.
Heine–Cantor Theorem — kai items mein test hone wala compact-domain upgrade.
Compactness / Lebesgue number lemma — kyun "finite subcover, min lo" kaam karta hai.
Lipschitz continuity — woh sufficient-not-necessary test.
Mean Value Theorem — bounded derivative ko Lipschitz se bridge karta hai.
Cauchy sequences — woh preservation property jo uniform ko pointwise se alag karti hai.