p.111 ex 2 : \(P_y(x)=\frac{1}{\pi}\cdot \frac{y}{x^2+y^2},\ x\in\mathbb{R},\ y>0.\) Show that \(P_y(x)\) is an approximation to identity \((\delta=y)\)

proof

① \(\int_{\mathbb{R}}P_y(x)\,dx=1\)

\(P_y(x)=\frac{1}{\pi}\cdot\frac{y}{x^2+y^2}=\frac{1}{\pi y}\cdot\frac{1}{1+(x/y)^2}\). Let \(\frac{x}{y}=t\ \to\ \frac{1}{y}dx=dt\)

\[\begin{aligned} \int_{\mathbb{R}}P_y(x)\,dx &=\frac{1}{\pi}\int_{\mathbb{R}}\frac{1}{1+t^2}\,dt =\frac{2}{\pi}\int_0^\infty \frac{1}{1+t^2}\,dt \quad (t=\tan\theta) \\ &=\frac{2}{\pi}\int_0^{\pi/2}\frac{1}{1+\tan^2\theta}\sec^2\theta\,d\theta =\frac{2}{\pi}\int_0^{\pi/2} d\theta =\frac{2}{\pi}\cdot\frac{\pi}{2}=1\ // \end{aligned}\]

② \(|P_y(x)|\le A\cdot y^{-1}\) \(::\) \(P_y(x)=\frac{1}{\pi y}\cdot\frac{1}{1+(x/y)^2}\le \frac{1}{\pi}\cdot\frac{1}{y}\)

③ \(|P_y(x)|\le A\cdot \frac{y}{x^2}\) \(::\) \(P_y(x)=\frac{1}{\pi}\cdot\frac{y}{x^2+y^2}\le \frac{1}{\pi}\cdot\frac{y}{x^2}\)

\(\therefore\ P_y(x)\) is an approximation to identity

p.112 ex5 : \[F_N(x)=\begin{cases}\frac{1}{2\pi N}\frac{\sin^2(Nx/2)}{\sin^2(x/2)} & |x|\le \pi \\0 & \text{o.w}\end{cases}\quad (\text{in }\mathbb{R}),\ \text{where }\delta=\frac{1}{N}\] Show that \(F_N\) is approximations to the identity.

proof

① \(\int F_N(x)\,dx=1\)

\(F_N(x)=\frac{1}{N}\sum_{n=0}^{N-1}D_n(x)\), where \(D_n(x)=\frac{1}{2\pi}\sum_{k=-n}^{n}e^{ikx}\ \ (x\in[-\pi,\pi])\)

\[\int_{-\pi}^{\pi}D_n(x)\,dx=1\ \ \left(\because \int_{-\pi}^{\pi}e^{inx}\,dx=\begin{cases}2\pi & (n=0)\\0 & (n\ne 0)\end{cases}\right),\ \ \text{so }\int_{-\pi}^{\pi}F_N(x)\,dx=1\]

③ \(|F_N(x)|\le A\cdot N\)

Since \(|D_n(x)|\le \frac{1}{2\pi}\sum_{k=-n}^{n}|e^{ikx}|=\frac{1}{2\pi}(2n+1)\), \[\begin{aligned} |F_N(x)| &\le \frac{1}{N}\sum_{n=0}^{N-1}|D_n(x)| \le \frac{1}{N}\cdot\frac{1}{2\pi}\sum_{n=0}^{N-1}(2n+1) \\ &=\frac{1}{2\pi N}\left\{2\cdot\frac{N(N-1)}{2}+N\right\} \\ &=\frac{1}{2\pi}(N-1+1)=\frac{1}{2\pi}\cdot N \end{aligned}\]

③ \(|F_N(x)|\le A\cdot \frac{1}{N}\cdot \frac{1}{x^2}\)

Note that \(\frac{|x|}{\pi}\le |\sin\frac{x}{2}|\). Then

\[\begin{aligned} |F_N(x)| &=\frac{1}{2\pi N}\cdot \frac{(\sin\frac{Nx}{2})^2}{(\sin\frac{x}{2})^2} \le \frac{1}{2\pi N}\cdot \frac{1}{(\sin\frac{x}{2})^2} \le \frac{1}{2\pi N}\cdot \frac{\pi^2}{x^2} = \frac{\pi}{2N}\cdot \frac{1}{x^2} \end{aligned}\] \(\therefore\ F_N(x)\) is an approximation to identity

Q1. \(F\in BV[a,b]\), then \(|F|\le M\)

proof \(|F(x)-F(a)|\le T_F(a,b)\) and \(|F(x)|-|F(a)|\le |F(x)-F(a)|\). So, \(|F(x)|\le |F(a)|+T_F(a,b)\quad \forall x\in[a,b]\).

Q2. Let \(F,G\in BV[a,b]\), and \(c\in\mathbb{R}\).

(i) \(F+G\), \(FG\), \(cF\in BV[a,b]\)

proof

① \(F+G\)

Let \(P=\{a=x_0<x_1<\cdots<x_n=b\}\) be a partition of \([a,b]\),

\[\begin{aligned} \sum_{i=1}^n |F(x_i)+G(x_i)-F(x_{i-1})-G(x_{i-1})| &\le \sum_{i=1}^n|F(x_i)-F(x_{i-1})|+\sum_{i=1}^n|G(x_i)-G(x_{i-1})| \\ &\le T_F(a,b)+T_G(a,b) \end{aligned}\]

② \(FG\) (let \(|F|\le L_F\) and \(|G|\le L_G\))

\[\begin{aligned} \sum_{i=1}^n |F(x_i)G(x_i)-F(x_{i-1})G(x_{i-1})| &=\sum_{i=1}^n |F(x_i)G(x_i)-F(x_{i-1})G(x_i)+F(x_{i-1})G(x_i)-F(x_{i-1})G(x_{i-1})| \\ &\le \sum_{i=1}^n |G(x_i)|\,|F(x_i)-F(x_{i-1})|+\sum_{i=1}^n|F(x_{i-1})|\,|G(x_i)-G(x_{i-1})| \\ &\le L_G\cdot M_F + L_F\cdot M_G \quad (\because F,G \text{ is bdd}) \end{aligned}\]

③ trivial

(ii) \(T_{F+G}(a,b)\le T_F(a,b)+T_G(a,b),\ \ T_{cF}(a,b)=c\cdot T_F(a,b)\)

proof ①

\[\sum_{i=1}^n |F(x_i)+G(x_i)-F(x_{i-1})-G(x_{i-1})| \le \sum_{i=1}^n|F(x_i)-F(x_{i-1})|+\sum_{i=1}^n|G(x_i)-G(x_{i-1})|.\]

This holds for all partitions, so \[T_{F+G}(a,b)\le \sum_{i=1}^n|F(x_i)-F(x_{i-1})|+\sum_{i=1}^n|G(x_i)-G(x_{i-1})|,\] and thus, \(T_{F+G}(a,b)\le T_F(a,b)+T_G(a,b)\)

②

\[\sum_{i=1}^n |cF(x_i)-cF(x_{i-1})|=c\sum_{i=1}^n|F(x_i)-F(x_{i-1})|.\]

\[\sup_P \sum_{i=1}^n |cF(x_i)-cF(x_{i-1})| = T_{cF}(a,b)\ge c\cdot \sum_{i=1}^n|F(x_i)-F(x_{i-1})| \Rightarrow T_{cF}(a,b)\ge c\cdot T_F(a,b).\]

\[\sum_{i=1}^n |cF(x_i)-cF(x_{i-1})| \le c\cdot \sup_P \sum_{i=1}^n|F(x_i)-F(x_{i-1})| = c\cdot T_F(a,b)\Rightarrow T_{cF}(a,b)\le c\cdot T_F(a,b).\]

\(\therefore\ T_{cF}(a,b)=c\cdot T_F(a,b)\)

Q3. \(F\in BV[a,b]\), \(c\in(a,b)\)

(i) \(F\in BV[a,c]\cap BV[c,b]\) \(V[F,P]\) : variation of \(F\) over \(P\)

proof Let \(P_1,P_2\) be partition of \([a,c]\) and \([c,b]\) respectively, and \(P=P_1\cup P_2\). Then, \(V[F,P_1]+V[F,P_2]=V[F,P]\le T_F(a,b)\).

\(V[F,P_1]\le T_F(a,b)\) for all \(P_1\), so \(F\in BV[a,c]\) \(V[F,P_2]\le T_F(a,b)\) for all \(P_2\), so \(F\in BV[c,b]\)

(ii) \(T_F(a,b)=T_F(a,c)+T_F(c,b)\)

proof ① \(T_F(a,c)+T_F(c,b)\le T_F(a,b)\) trivial

From (i), \(V[F,P_1]+V[F,P_2]\le T_F(a,b)\) for all \(P_1\) & \(P_2\). So, \(T_F(a,c)+T_F(c,b)\le T_F(a,b)\).

② \(T_F(a,b) \le T_F(a,c) + T_F(c,b)\)

Let \(Q\) be a partition of \([a,b]\), \(\hat{Q}=Q\cup\{c\}\), \(Q_1=\hat{Q}\cap[a,c]\), and \(Q_2=\hat{Q}\cap[c,b]\).

\(V(F,Q)\le V(F,\hat{Q})=V(F,Q_1)+V(F,Q_2)\le T_F(a,c)+T_F(c,b)\).

This holds for all \(Q\), so \(T_F(a,b)\le T_F(a,c)+T_F(c,b)\).

Q4. \(F:[a,b]\to\mathbb{R}\) is continuous and \(F\in BV[a,b]\). Then \(T_F(a,x)\) is continuous \((x\in(a,b))\)

proof Consider \(c\in(a,b)\). \(F\) is continuous at \(c\), so there is \(\delta_0>0\) such that \(|x-c|<\delta_0\) implies \(|F(x)-F(c)|<\varepsilon/2\) for given \(\varepsilon>0\) \((x\in[a,b])\).

Since \(F\in BV[a,b]\), there is a partition of \([a,b]\) \(P_0\) such that \(T_F(a,b) < V(F,P_0) + \varepsilon/2\).

\(\delta_0 > |b-a|\) \(\delta_0 \le b-a\)

Let \(P = P_0 \cup \{ \{c-\delta_0/2,\ c,\ c+\delta_0/2\}\cap[a,b] \} = \{\hat x_0,\cdots,x_{k_1},c,x_{k_1+1},\cdots,\hat x_N\}\) and \(P_1=\{\hat x_0,\cdots,x_{k_1}\}\), \(P_2=\{x_{k_1+1},\cdots,\hat x_N\}\).

\(|c-x_{k_1}|,\ |c-x_{k_1+1}| \le \delta_0/2 < \delta_0\), then \[ \begin{aligned} V(F,P) &= V(F,P_1)+V(F,P_2)+|F(c)-F(x_{k_1})|+|F(x_{k_1+1})-F(c)| \\ &< V(F,P_1)+V(F,P_2)+\varepsilon/2+\varepsilon/2 \\ &\le T_F(a,x_{k_1}) + T_F(x_{k_1+1},b) + \varepsilon . \end{aligned} \]

Then, \(T_F(a,b) \le T_F(a,x_{k_1}) + T_F(x_{k_1+1},b) + \varepsilon\) and \(T_F(a,x_{k_1}) \le T_F(a,x_{k_1}) + \varepsilon\).

Let \(\delta_\varepsilon=\min(c-x_{k_1},x_{k_1+1}-c)\), then \(|x-c|<\delta_\varepsilon\) implies \(x\in[x_{k_1},x_{k_1+1}]\) and thus \(|T_F(a,x)-T_F(a,c)| \le T_F(a,x_{k_1+1})-T_F(a,x_{k_1}) < \varepsilon_\varepsilon\).

This holds for all \(\varepsilon>0\), so \(T_F(a,x)\) is cont at \(c\in(a,b)\) and thus at all \(x\in(a,b)\)

Q5. \(F:[a,b]\to\mathbb{R}\) is a continuous function. Then \(F\in BV[a,b]\) if \(F\) is a difference of two increasing and continuous function.

proof ① \(\Leftarrow\) (Thm 3.2.3)
\(F=F_1-F_2\), where \(F_1,F_2\) are increasing & continuous. Then \(F\in BV[a,b]\).

② \(\Rightarrow\)
Then, \(T_F(a,x)\) is continuous. Let \(P_F(a,x)=\sup\left\{\sum_{i=1}^k \big(F(t_i)-F(t_{i-1})\big): F(t_i)-F(t_{i-1})\ge 0\right\}\) and \(N_F(a,x)=\sup\left\{\sum_{i=1}^k -\big(F(t_i)-F(t_{i-1})\big): F(t_i)-F(t_{i-1})\le 0\right\}\), which are increasing. Then \[ \begin{aligned} F(x)-F(a) &= P_F(a,x)-N_F(a,x),\\ T_F(a,x) &= P_F(a,x)+N_F(a,x). \end{aligned} \]

Since \(P_F(a,x)=\frac12\big(F(x)+T_F(a,x)-F(a)\big)\) and \(N_F(a,x)=\frac12\big(T_F(a,x)-F(x)+F(a)\big)\), both \(P_F\) and \(N_F\) are continuous.

Thus, \(F=(P_F+F(a))-N_F\) is a difference of increasing & continuous functions.