Exercises — Uniform convergence of function sequences

Move 1 — pointwise limit. $x\in [0,1]$ fix karo. Jab $n\to \infty$, $\frac{x}{n}\to 0$. Toh $f(x)=0$ har $x$ ke liye. (Koi case-splitting nahi chahiye — sabhi $x$ ke liye same hota hai.)

Move 2 — worst gap. Gap hai $|f_n(x)-f(x)|=\frac{x}{n}$. $[0,1]$ par yeh $x$ ke saath badhta hai, isliye $x=1$ par sabse bada hai.

Move 3 — sup. $M_n={\sup }_{x\in [0,1]}\frac{x}{n}=\frac{1}{n}\to 0.$ Kyunki $M_n\to 0$, convergence $[0,1]$ par uniform hai. $■$

Move 1. Koi bhi $x\ge 0$ fix karo: $\frac{x}{n}\to 0$, toh pointwise abhi bhi $f(x)=0$ hai.

Move 2. Gap $\frac{x}{n}$ ka ab koi largest point nahi — jab $x\to \infty$ toh yeh unbounded grow karta hai.

Move 3. $M_n={\sup }_{x\ge 0}\frac{x}{n}=+\infty \text{for every }n.$ $M_n\nrightarrow 0$ (yeh kabhi finite bhi nahi hai), isliye convergence $[0,\infty )$ par uniform nahi hai.

Moral: same formula, alag domain, alag answer — uniformity $(f_n,E)$ dono ki saath milke ek property hai, sirf formula ki nahi. $■$

Move 1 — pointwise. $x\in [0,1)$ ke liye, $x^n\to 0$ isliye $x^n(1-x)\to 0$. $x=1$ par, factor $(1-x)=0$ isliye $f_n(1)=0$. Toh $f(x)=0$ har jagah.

Move 2 — moving peak. Graph upar jaata hai phir neeche aata hai; peak $n$ badhne ke saath $1$ ki taraf move karta hai. Use dhundne ke liye differentiate karo: $f_n^{\prime }(x)=n{x}^{n-1}(1-x)-x^n=x^{n-1}(n-(n+1)x).$ Bracket ko $0$ set karne par interior maximum milta hai $x_n=\frac{n}{n+1}$ par.

Move 3 — sup value. $M_n=f_n(x_n)=\left(\frac{n}{n+1}{)}^n\right(1-\frac{n}{n+1})=(\frac{n}{n+1}{)}^n\cdot \frac{1}{n+1}.$ Ab $(\frac{n}{n+1}{)}^n=(1+\frac{1}{n}{)}^{-n}\to e^{-1}$, aur $\frac{1}{n+1}\to 0$. Isliye $M_n\to e^{-1}\cdot 0=0.$ Toh convergence $[0,1]$ par uniform hai. $■$

Extra factor $(1-x)$ troublesome peak height ko $0$ par crush kar deta hai — plain $x^n$ se contrast karo jahan peak height $1$ par tiki rehti thi.

Move 1 — pointwise. $x=0$ par: $f_n(0)=0$. Fixed $x>0$ ke liye: numerator $\sim nx$, denominator $\sim n^2{x}^2$, isliye $f_n(x)\sim \frac{1}{nx}\to 0$. Toh $f(x)=0$ har jagah.

Move 2 — moving peak. $u=nx$ substitute karo: $f_n=\frac{u}{1+u^2}$, jo $u=1$ par peak karta hai, yaani $x_n=\frac{1}{n}\in [0,1]$ sabhi $n\ge 1$ ke liye.

Move 3 — sup value. $x_n=1/n$ par: $M_n=\frac{n\cdot \frac{1}{n}}{1+n^2\cdot \frac{1}{n^2}}=\frac{1}{1+1}=\frac{1}{2}\text{for every }n.$ $M_n=\frac{1}{2}\nrightarrow 0$, isliye convergence $[0,1]$ par uniform nahi hai. Fixed height $\frac{1}{2}$ ka ek spike $0$ ki taraf slide karta hai aur kabhi shrink nahi karta. $■$

Move 1 — pointwise. $x=0$ par: $f_n(0)=0$. Fixed $x>0$ ke liye: jab $n\to \infty$, $(1-nx{)}^2\to \infty$, isliye denominator blow up hota hai aur $f_n(x)\to 0$. Toh $f(x)=0$.

Move 2 — clever bad point. Denominator tab sabse chhota hai jab $1-nx=0$, yaani $x=\frac{1}{n}$. $x_n=\frac{1}{n}$ choose karo (jo $[0,1]$ mein hai).

Move 3 — evaluate. $f_n\left(\frac{1}{n}\right)=\frac{(1/n{)}^2}{(1/n{)}^2+0}=1.$ Isliye $M_n\ge f_n(1/n)=1$ sabhi $n$ ke liye, toh $M_n\nrightarrow 0$: uniform nahi.

Note karo ki pointwise convergence har fixed $x$ par hold karta hai (freeze karo $x$, point $1/n$ eventually use pass ho jaata hai), lekin moving point $x_n=1/n$ par value $1$ par pinned rehti hai. Yahi reason hai ki uniformity strictly stronger kyun hai. $■$

Har $f_n(x)=x^n$ $[0,1]$ par continuous hai. Pointwise limit hai $f(x)=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array},$ jo $x=1$ par discontinuous hai.

Theorem (contrapositive): agar continuous functions ki ek sequence uniformly converge karti hai, toh limit continuous hoti hai. Yahan limit continuous nahi hai, isliye convergence uniform nahi ho sakti. $■$

Yeh ek "crime ko uske footprint se detect karo" argument hai (dekho Continuity preserved under uniform limits): continuous functions ki discontinuous limit non-uniformity ka proof hai, koi sup computation nahi chahiye.

Move 1 — pointwise limit on $[0,\frac{1}{2}]$. Har $x\in [0,\frac{1}{2}]$ ke liye $x<1$ hai, isliye $x^n\to 0$; toh $f\equiv 0$ yahan.

Move 2 — uniformity. $M_n={\sup }_{x\in [0,1/2]}{x}^n={\left(\frac{1}{2}\right)}^n\to 0$, isliye convergence uniform hai. Interchange of limit and integral theorem ke zariye hum $\lim$ aur $\int$ swap kar sakte hain.

Move 3 — dono sides explicitly check karo.

• Right side: ${\int }_0^{1/2}0dx=0.$
• Left side: ${\int }_0^{1/2}{x}^{n}dx=\frac{(1/2{)}^{n+1}}{n+1}$, aur yeh $\to 0$ jab $n\to \infty$.

Dono $0$ ke barabar hain. Interchange valid hai, exactly jaise theorem ne promise kiya tha. $■$

Move 1 — pointwise limit. $x=0$ aur $x=1$ par, $g_n=0$. Fixed $x\in (0,1)$ ke liye: $(1-x{)}^n\to 0$ geometrically, jo polynomial $n^2$ ko beat karta hai, isliye $g_n(x)\to 0$. Toh $g\equiv 0$ pointwise.

Move 2 — the integral. ${\int }_0^{1}x(1-x{)}^{n}dx=\frac{1}{(n+1)(n+2)}$ use karo (Beta integral). Isliye ${\int }_0^1{g}_{n}dx=n^2\cdot \frac{1}{(n+1)(n+2)}=\frac{n^2}{(n+1)(n+2)}\to 1.$

Move 3 — interpret. Pointwise-limit integral $0$ hai, lekin ${\lim }_n\int g_n=1\ne 0$. Interchange fail hua. Interchange Theorem ke contrapositive se, $g_n$ $[0,1]$ par uniformly converge nahi karta. Waakai ek tall thin spike (height $\sim n^2\cdot \frac{1}{n}=n$ near $x\sim \frac{1}{n}$) ek fixed amount of area finish line tak le jaata hai. $■$

Dominated convergence theorem se contrast karo: woh softer tool bhi yahan fail ho jaata kyunki koi integrable function nahi hai jo sabhi $g_n$ ko dominate kare.

Move 1 — tail ko uniformly bound karo. $m>n$ ke liye, $|f_m(x)-f_n(x)|=\left|{\sum }_{k=n+1}^m\frac{\cos (kx)}{k^2}\right|\le {\sum }_{k=n+1}^m\frac{|\cos (kx)|}{k^2}\le {\sum }_{k=n+1}^m\frac{1}{k^2}.$ Bound $\frac{1}{k^2}$ $x$ se independent hai — yahi Weierstrass $M$-test in action hai, $M_k=\frac{1}{k^2}$ ke saath aur $\sum M_k=\frac{{\pi }^2}{6}<\infty$.

Move 2 — sup lo, phir $n\to \infty$ bhejo. Kyunki bound mein koi $x$ nahi, ${\sup }_{x\in \mathbb{R}}|f_m(x)-f_n(x)|\le {\sum }_{k=n+1}^{\infty }\frac{1}{k^2}=:R_n.$ Kyunki $\sum 1/k^2$ converge karta hai, uska tail $R_n\to 0$. Toh diya hua $\epsilon >0$ ke liye hum $N$ pick karte hain $R_N<\epsilon$ ke saath; phir sabhi $m,n\ge N$ ke liye sup-distance $<\epsilon$ hai.

Move 3 — conclude. Yahi exactly uniform Cauchy criterion hai, isliye $(f_n)$ $\mathbb{R}$ par uniformly converge karta hai — even though humne limit function kabhi likhi nahi. (Dekho Cauchy sequences in metric spaces: sup-norm ke saath $C(\mathbb{R})$ complete hai, isliye uniform-Cauchy $\Rightarrow$ uniform-convergent.) $■$

Move 1 — pointwise. $x=0$ par, $f_n(0)=0$. Fixed $x>0$ ke liye, exponential $e^{-n{x}^2}\to 0$ linear factor $nx$ ko beat karta hai, isliye $f_n(x)\to 0$. Toh $f\equiv 0$.

Move 2 — moving peak. $x$ mein differentiate karo: $f_n^{\prime }(x)=n{e}^{-n{x}^2}(1-2n{x}^2)=0⟹x_n=\frac{1}{\sqrt{2n}}.$ Peak height: $M_n=f_n(x_n)=n\cdot \frac{1}{\sqrt{2n}}{e}^{-1/2}=\sqrt{\frac{n}{2}}{e}^{-1/2}\to \infty .$ $M_n\to \infty \nrightarrow 0$: convergence uniform nahi hai (ek spike jo bhi badhta hai aur narrow bhi hota hai).

Move 3 — the integrals. $u=n{x}^2$, $du=2nxdx$ substitute karo: ${\int }_0^{\infty }nx{e}^{-n{x}^2}dx=\frac{1}{2}{\int }_0^{\infty }{e}^{-u}du=\frac{1}{2}\text{for every }n.$ Toh $\underset{n}{\lim }{\int }_0^{\infty }{f}_{n}dx=\frac{1}{2}$, lekin ${\int }_0^{\infty }\underset{n}{\lim }{f}_{n}dx={\int }_0^{\infty }0dx=0$.

Kyunki $\frac{1}{2}\ne 0$, interchange fail hua — Move 2 mein milī non-uniformity se consistent. Fixed area $\frac{1}{2}$ shrinking spike par sawaar hokar jaata hai. $■$

• $x/n$ on $[0,1]$ vs $[0,\infty )$? ::: $M_n=1/n\to 0$ (uniform) vs $M_n=\infty$ (not uniform).
• $x^n(1-x)$ on $[0,1]$? ::: $M_n=(n/(n+1){)}^n/(n+1)\to 0$, uniform.
• $nx/(1+n^2{x}^2)$ on $[0,1]$? ::: Peak at $x_n=1/n$, $M_n=1/2$, not uniform.
• Fastest non-uniformity proof for $x^n$ on $[0,1]$? ::: Limit is discontinuous at $1$; contrapositive of continuity theorem.
• $nx{e}^{-n{x}^2}$ on $[0,\infty )$: is $\lim \int =\int \lim$? ::: No: $1/2\ne 0$; spike carries fixed area.
• Tool to prove uniformity without the limit? ::: Uniform Cauchy criterion / Weierstrass $M$-test.