Exercises — Uniform continuity — difference from pointwise

$\forall \epsilon >0\exists \delta >0\forall x,y\in A:|x-y|<\delta ⟹|f(x)-f(y)|<\epsilon .$ Ek hi difference: yahan $\exists \delta$ points $\forall x,y$ se pehle aata hai, isliye $\delta =\delta (\epsilon )$ sirf. Pointwise continuity mein $\exists \delta$ point $\forall x_0$ ke baad aata hai, isliye $\delta =\delta (\epsilon ,x_0)$ point ko dekh sakta hai. ==One $\delta$ for everyone== — yahi poori kahani hai.

$|x^2-y^2|=|x+y||x-y|.$ Amplifying factor hai $==|x+y|==$. $\mathbb{R}$ par yeh unbounded hai, isliye wohi input gap $|x-y|$ door jaane par ek aur bada output gap produce karta hai. Neeche figure dekho — required input width shrink hoti jaati hai jab $x$ badhta hai.

(a) T — $\delta =\epsilon /L$ lo. (b) F — $\sqrt{x}$ on $[0,\infty )$ uniformly continuous hai lekin Lipschitz nahi (iska slope $0$ ke paas blow up karta hai). (c) T — Heine–Cantor. (d) F — $(0,1)$ closed/compact nahi hai; $1/x$ ek counterexample hai.

$|f(x)-f(y)|=|3x-3y|=3|x-y|$, toh $f$ Lipschitz hai $L=3$ ke saath. Uniform continuity: $\delta =\epsilon /3$ lo. $\epsilon =0.03$ ke liye, $\delta =0.01$. Check: $|x-y|<0.01\Rightarrow |f(x)-f(y)|=3|x-y|<0.03$. $\delta$ mein koi $x$ nahi hai — yahi uniformity hai.

$x,y\in [0,5]$ ke liye: $|x+y|\le 10$, toh $|x^2-y^2|=|x+y||x-y|\le 10|x-y|.$ Is tarah $f$ $[0,5]$ par Lipschitz hai $L=10$ ke saath; $\delta =\epsilon /10$ lo. $\epsilon =0.1$ ke liye, $\delta =0.01$. Bounded factor $|x+y|\le 10$ exactly wahi hai jo $\mathbb{R}$ mein nahi tha — yeh Heine–Cantor ka action hai compact interval par.

Di gayi inequality se, $\delta ={\epsilon }^2$ choose karo. Tab $|x-y|<{\epsilon }^2⟹|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}<\sqrt{{\epsilon }^2}=\epsilon .$ $\epsilon =0.02$ ke liye, $\delta =0.0004$. Note karo ki $f^{\prime }(x)=\frac{1}{2\sqrt{x}}\to \infty$ near $0$, isliye $f$ Lipschitz nahi hai — phir bhi uniformly continuous hai. Lipschitz sufficient hai, necessary nahi.

Inequality kyun hold hoti hai: $(\sqrt{x}-\sqrt{y}{)}^2=x+y-2\sqrt{xy}\le x+y-2\min (x,y)=|x-y|$, kyunki $\sqrt{xy}\ge \min (x,y)$.

Fix karo $\epsilon =1$. Koi bhi $\delta \in (0,1)$ diya ho, ek chhota $x$ aur uska neighbour lo: $x=\delta ,y=\frac{\delta }{2}.$ Dono $(0,1)$ mein hain, aur $|x-y|=\delta /2<\delta$. Lekin $|\frac{1}{x}-\frac{1}{y}|=|\frac{1}{\delta }-\frac{2}{\delta }|=\frac{1}{\delta }>1=\epsilon (\text{since }\delta <1).$ Input mein close, output mein door — formal negation. Problem boundary point $0$ hai (excluded), jahan slope $-1/x^2$ blow up karta hai. Figure dekho: do nearby inputs, huge output gap.

Continuity: $(0,1)$ par continuous maps ka composition (jahan $x\ne 0$). Ab $\epsilon =1$ ke saath uniformity todo. Wo points choose karo jahan sine opposite peaks hit kare: $x_n=\frac{1}{2\pi n+\frac{\pi }{2}},y_n=\frac{1}{2\pi n-\frac{\pi }{2}}.$ Tab $\sin (1/x_n)=\sin (2\pi n+\frac{\pi }{2})=1$ aur $\sin (1/y_n)=\sin (2\pi n-\frac{\pi }{2})=-1$, toh $|f(x_n)-f(y_n)|=|1-(-1)|=2>\epsilon .$ Phir bhi $|x_n-y_n|=\frac{\pi }{(2\pi n{)}^2-(\pi /2{)}^2}\to 0$ as $n\to \infty$. Toh koi bhi $\delta$ ke liye $n$ itna bada lo ki $|x_n-y_n|<\delta$ — output gap $2$ rehta hai. Uniformly continuous nahi.

Maano $a_n=1/n\in (0,1)$. Yeh Cauchy hai (yeh $0$ ki taraf converge karta hai, toh iske terms ek saath bunch ho jaate hain). $f$ apply karo: $f(a_n)=n$. Image $(1,2,3,\dots )$ Cauchy nahi hai (terms hamesha ke liye $1$ se differ karte hain). Ek uniformly continuous function zaroor Cauchy sequences ko Cauchy sequences mein bhejna chahiye, toh $1/x$ $(0,1)$ par uniformly continuous nahi ho sakta. Yeh ek slick criterion hai: Cauchy sequence ko todta hai $\Rightarrow$ uniformly continuous nahi.

Fix karo $\epsilon >0$. Uniform continuity se ${\delta }_f$ milta hai tolerance $\epsilon /2$ ke liye ($f$ par) aur ${\delta }_g$ milta hai tolerance $\epsilon /2$ ke liye ($g$ par). Maano $\delta =\min ({\delta }_f,{\delta }_g)>0$ — yeh ab bhi sirf $\epsilon$ par depend karta hai. Tab $|x-y|<\delta$ ke liye, $|(f+g)(x)-(f+g)(y)|\le |f(x)-f(y)|+|g(x)-g(y)|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .$ Triangle inequality + $\epsilon$ ko half mein split karna — yahi workhorse hai. $\delta$ ne $x$ kabhi nahi dekha, toh $f+g$ uniformly continuous hai.

$f(x)=x$: $|x-y|\le 1\cdot |x-y|$, Lipschitz with $L=1$, toh $\delta =\epsilon$ ke saath uniformly continuous. Lekin $f\cdot f=x^2$, jo parent note (aur figure s01) dikhata hai $\mathbb{R}$ par fail hota hai: fix karo $\epsilon =1$; kisi bhi $\delta$ ke liye $x=\frac{1}{\delta },y=\frac{1}{\delta }+\frac{\delta }{2}$ lo, deta hai $|x^2-y^2|=(\frac{2}{\delta }+\frac{\delta }{2})\frac{\delta }{2}=1+\frac{{\delta }^2}{4}>1.$ Toh product ne uniformity tod di jab ki har factor ne use rakha. (Rescue: kisi bhi bounded domain par uniformly continuous functions ka product uniformly continuous rehta hai, kyunki functions tab bounded hote hain.)

Fix karo $\epsilon >0$. $g$ ki uniform continuity $\eta >0$ deti hai jiske saath $|u-v|<\eta \Rightarrow |g(u)-g(v)|<\epsilon$ (for $u,v\in B$). $\eta$ ko target tolerance ki tarah $f$ mein daalo: $f$ ki uniform continuity $\delta >0$ deti hai jiske saath $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\eta$. Inhe chain karo: $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\eta \Rightarrow |g(f(x))-g(f(y))|<\epsilon$. Dono $\delta ,\eta$ sirf $\epsilon$ par depend karte hain — uniformity composition survive kar leti hai.

($\Leftarrow$) Agar $\omega$ exist karta hai $\omega (t)\to 0$ ke saath, toh $\epsilon >0$ diya ho, $\delta >0$ choose karo jisme $\omega (t)<\epsilon$ for all $t<\delta$ (possible kyunki $\omega (t)\to 0$). Tab $|x-y|<\delta \Rightarrow |f(x)-f(y)|\le \omega (|x-y|)<\epsilon$. Uniform. ($\Rightarrow$) Define karo $\omega (t)=\sup \{|f(x)-f(y)|:|x-y|\le t\}$. Uniform continuity kehti hai: har $\epsilon$ ke liye ek $\delta$ hai jo har aisi difference ko $<\epsilon$ banata hai jab $t<\delta$ ho, matlab $\omega (t)\le \epsilon$ for $t<\delta$. Isliye $\omega (t)\to 0$ as $t\to 0^{+}$, aur construction se $|f(x)-f(y)|\le \omega (|x-y|)$. $\sqrt{x}$ ke liye: $\omega (t)=\sqrt{t}\to 0$. $\sqrt{t}<\epsilon$ solve karne se $t<{\epsilon }^2$ milta hai, L2.3 se $\delta ={\epsilon }^2$ recover hota hai.

Koi bhi sequence $x_n\to a^{+}$ inside $(a,b)$ lo; yeh Cauchy hai. L3.3 ke principle se (uniform continuity Cauchyness preserve karti hai), $(f(x_n))$ $\mathbb{R}$ mein Cauchy hai, isliye converge karta hai (real Cauchy sequences converge karti hain). Yeh dekhne ke liye ki limit har aisi sequence ke liye same hai, do sequences $x_n\to a$, $x_n^{\prime }\to a$ ko ek sequence $x_1,x_1^{\prime },x_2,x_2^{\prime },\cdots \to a$ mein interleave karo: yeh Cauchy hai, iska image Cauchy hai, toh yeh ek value par converge karta hai — isse $\lim f(x_n)=\lim f(x_n^{\prime })$ force hota hai. Woh common value hi ${\lim }_{x\to a^{+}}f(x)$ hai. $f(a)$ ko is limit se define karna (aur similarly $f(b)$) $f$ ko closed, compact $[a,b]$ tak continuously extend karta hai, jis par Heine–Cantor uniform continuity confirm karta hai.

Contrast karo $1/x$ on $(0,1)$ se: yeh uniformly continuous nahi hai, aur indeed ${\lim }_{x\to 0^{+}}1/x=+\infty$ finitely exist nahi karta — koi continuous extension nahi. Consistent.

Answer: uniformly continuous iff $0<p\le 1$.

• $0<p\le 1$ (uniform): Yahan $x^p$ concave hai, aur $x,y\ge 0$ ke liye sub-additive bound $|x^p-y^p|\le |x-y{|}^p$ dikhaya ja sakta hai. Kyunki $t\mapsto t^p$ ek valid modulus of continuity hai ($t^p\to 0$ as $t\to 0$), L5.1 uniform continuity deta hai; explicitly $\delta ={\epsilon }^{1/p}$. ($p=1$ ke liye, $x^p=x$ Lipschitz hai.)
• $p>1$ (fails): slope $p{x}^{p-1}\to \infty$, aur $x^2$ ki tarah hum ek break engineer karte hain. Fix karo $\epsilon =1$. Kisi bhi $\delta >0$ ke liye, $x$ bada lo aur $y=x+\frac{\delta }{2}$. MVT se, $x^p-y^p$... rather $y^p-x^p=p{c}^{p-1}\cdot \frac{\delta }{2}$ kisi $c\in (x,y)$ ke liye; jab $x\to \infty$, $c^{p-1}\to \infty$, toh output gap $\to \infty >1$ jab input gap $=\delta /2<\delta$ ho. Koi fixed $\delta$ kaam nahi karta. (Parent ka $x^2$ case $p=2$ hai.)
• Boundary $p=1$: exactly Lipschitz, crossover point — uniform.

Toh threshold precisely $p=1$ par hai: sub-linear/linear growth (bounded ya tapering slope effect $t^p$ modulus se) safe hai; super-linear growth ($p>1$) slope ko bhaagne deta hai.