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Solution to Principles of Mathematical Analysis Chapter 6 Part B


Chapter 6 The Riemann-Stieltjes Integral

Exercise 11

(By analambanomenos) As in the proof of Theorem 1.37(f), we show that
\begin{align*}
||u+v||_2^2 &=\int_a^b|u+v|^2\,d\alpha \\
&\le \int_a^b\bigl(|u|^2+2|uv|+|v|^2\bigr)\,d\alpha \\
&= ||u||_2^2+2\int_a^b|uv|\,d\alpha+||v||_2^2 \\
&\le ||u||_2^2+2||u||_2||v||_2+||v||_2^2 \\
&= \bigl(||u||_2+||v||_2\bigr)^2
\end{align*} so that $||u+v||_2\le||u||_2+||v||_2$. The result follows by letting $u=f-g$ and $v=g-h$.


Exercise 12

(By analambanomenos) Since $f\in\mathscr R(\alpha)$, There are bounds $m\le f(x)\le M$ for the values of $f$ in $[a,b]$. Let $\varepsilon>0$ and let $P=\{x_0,\ldots,x_n\}$ be a partition of $[a,b]$ such that $$U(f,P,\alpha)-L(f,P,\alpha)\le\varepsilon^2/(M-m).$$ Following the hint, for $t\in[x_{i-1},x_i]$ define $$g(t)=\frac{x_i-t}{\Delta x_i}f(x_{i-1})+
\frac{t-x_{i-1}}{\Delta x_i}f(x_i).$$ Then $g$ is linear on $[x_{i-1},x_i]$ and the definitions of $g$ on $[x_{i-1},x_i]$ and $[x_i,x_{i+1}]$ both define $g(x_i)=f(x_i)$, and so $g$ is a continuous function on $[a,b]$. Also, $m_i\le g(t)\le M_i$ on each $[x_{i-1},x_i]$. Hence
\begin{align*}
||f-g||_2^2 &= \int_a^b|f-g|^2\,d\alpha \\
&= \sum_{i=0}^n\int_{x_{i-1}}^{x_i}|f-g|^2\,d\alpha \\
&\le \sum_{i=0}^n(M_i-m_i)^2\Delta\alpha_i \\
&\le (M-m)\sum_{i=0}^n(M_i-m_i)\Delta\alpha_i \\
&= (M-m)\bigl(U(f,P,\alpha)-L(f,P,\alpha)\bigr) \\
&\le \varepsilon^2
\end{align*}


Exercise 13

(By analambanomenos)

(a) Following the hint (substituting $u$ for $t^2$, integration by parts using $\sin u$ and $1/\bigl(2\sqrt{u}\bigr)$, replacing $|\cos u|$ with 1), we get
\begin{align*}
f(x) &= \int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt{u}}\,du \\
&= -\frac{\cos(x+1)^2}{2(x+1)}+\frac{\cos x^2}{2x}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\,du \\
\big|f(x)\big| &\le \frac{\big|\cos(x+1)^2\big|}{2(x+1)}+\frac{|\cos x^2|}{2x}+\int_{x^2}^{(x+1)^2}\frac{|\cos u|}{4u^{3/2}}\,du \\
&< \frac{1}{2(x+1)}+\frac{1}{2x}+\int_{x^2}^{(x+1)^2}\frac{1}{4u^{3/2}}\,du \\
&= \frac{1}{2(x+1)}+\frac{1}{2x}+\bigg(\frac{1}{2x}-\frac{1}{2(x+1)}\bigg) \\
&= \frac{1}{x}
\end{align*}(The strict inequality is due to the fact that $|\cos u|$ will not be constantly equal to its maximum possible value 1 throughout the interval of integration.)

(b) From the results in part (a), we get
\begin{align*}
2xf(x) &< \cos x^2 -\frac{x\cos(x+1)^2}{x+1}+\frac{1}{x+1} \\
r(x)=2xf(x)-\cos x^2+\cos(x+1)^2 &< \frac{\cos(x+1)^2}{x+1}+\frac{1}{x+1} \\
\big|r(x)\big| &< \frac{2}{x+1} < \frac{2}{x}
\end{align*}(c) (partial) From part (b), we get that as $x\rightarrow\infty$, $$xf(x)\approx\frac{\cos x^2-\cos(x+1)^2}{2}.$$ Hence $xf(x)$ lies (more or less) between $(-1-1)/2=-1$ and $(1+1)/2=1$.

(d) Note that $\sin t^2$ is positive between $\sqrt{n\pi}$ and $\sqrt{(n+1)\pi}$ for any even integer $n$ and negative for any odd integer $n$. Hence $\int_0^\infty\sin t^2\,dt$ can be reduced to an alternating series, whose terms satisfy $$\Bigg|\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}}\sin t^2\,dt\Bigg|<\sqrt{n\pi}-\sqrt{(n+1)\pi}=\frac{\sqrt{\pi}}{\sqrt{n}+\sqrt{n+1}}$$which goes to 0 as $n\rightarrow\infty$. Hence, by Theorem 3.43, the integral converges.


Exercise 14

(By analambanomenos) Substituting $u$ for $e^t$ and integrating by parts, we get
\begin{align*}
f(x) &= \int_x^{x+1}\sin e^t\,dt \\
&= \int_{e^x}^{e^{x+1}}\frac{\sin u}{u}\,du \\
&= \frac{\cos e^x}{e^x}-\frac{\cos e^{x+1}}{e^{x+1}}-\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}\,du
\end{align*}Hence, substituting $-1$ for $\cos u$ in the integral, we get
\begin{align*}
e^xf(x) &< \cos e^x-\frac{\cos e^{x+1}}{e}-e^x\int_{e^x}^{e^{x+1}}u^{-2}\,du \\
&= (1+\cos e^x) – e^{-1}(1+\cos e^{x+1}),
\end{align*}so that $e^x\big|f(x)\big|<2(1+1/e)$, which isn’t quite 2 (more work would need to be done to get this below 2). Doing another integration by parts, we get
\begin{align*}
r(x) &= -e^x\int_{e^x}^{e^{x+1}}\frac{\cos u}{u^2}\,du \\
&= e^x\Biggl(\frac{\sin e^x}{e^{2x}}-\frac{\sin e^{x+1}}{e^{2x+2}}-2\int_{e^x}^{e^{x+1}}\frac{\sin u}{u^3}\,du\Biggr) \\
&< \frac{\sin e^x}{e^x}-\frac{\sin e^{x+1}}{e^{x+2}}+2e^x\int_{e^x}^{e^{x+1}}u^{-3}\,du \\
&= \frac{1+\sin e^x}{e^x}-\frac{1+\sin e^{x+1}}{e^{x+2}},
\end{align*}so that $\big|r(x)\big|<2(1+1/e^2)e^{-x}$.


Exercise 15

(By analambanomenos) Applying integration by parts, applied to the functions $f^2$ and $1$, we get $$1=\int_a^bf^2(x)\,dx=bf^2(b)-af^2(a)-\int_a^b2xf(x)f’(x)\,dx=-2\int_a^bxf(x)f’(x)\,dx$$ which yields the first assertion. In the solution to Exercise 10 I showed that $$\biggl(\int_a^bF^2\,dx\biggr)^{1/2}\biggl(\int_a^bG^2\,dx\biggr)^{1/2}\ge\int_a^b|FG|\,dx$$ with equality only if $F^2$ is a multiple of $G^2$. Letting $F(x)=f’(x)$ and $G(x)=xf(x)$, then $$\biggl(\int_a^bf’^2(x)\,dx\biggr)^{1/2}\biggl(\int_a^bx^2f^2(x)\,dx\biggr)^{1/2}\ge
\int_a^b\big|xf(x)f’(x)\big|\,dx\ge\bigg|\int_a^bxf(x)f’(x)\,dx\bigg|=\frac{1}{2}.$$ Squaring both sides almost yields the second assertion, with $\ge$ instead of $>$. To get the strict inequality, note that the equality only holds if $f’^2(x)$ is a multiple of $x^2f^2(x)$. Since $\int_a^bf^2(x)\,dx=1$, the continuous function $f(x)$ cannot have the constant value 0. Since $f(a)=f(b)=0$, there are points $c$, $d$ such that $a\le c<d\le b$, $f(c)=f(d)=0$ and $f(x)\neq0$ for $c<x<d$. Then by Theorem 5.10 there is a point $x$ between $c$ and $d$ such that $f’(x)=0$. If equality were to hold above, then we would have $f’^2(x)=0=x^2f^2(x)\neq 0$, which is impossible.


Exercise 16

(By analambanomenos)

(a) We have
\begin{align*}
s\int_1^N\frac{[x]}{x^{s+1}}\,dx &= s\sum_{n=1}^{N-1}n\int_n^{n+1}x^{-s-1}\,dx \\
&= \sum_{n=1}^{N-1}n\biggl(\frac{1}{n^s}-\frac{1}{(n+1)^s}\biggr) \\
&= 1\biggl(\frac{1}{1^s}\biggr)+\Biggl(\sum_{n=2}^{N-1}\bigl(n-(n-1)\bigr)\frac{1}{n^s}\Biggr)-\frac{N-1}{N^s} \\
&= \Biggl(\sum_{n=1}^N\frac{1}{n^s}\Biggr)-\frac{1}{N^s}-\frac{N-1}{N^s} \\
&= \Biggl(\sum_{n=1}^N\frac{1}{n^s}\Biggr)-\frac{1}{N^{s-1}}
\end{align*}Taking the limit as $N\rightarrow\infty$, we get the desired result.

(b) From the work done in part (a), we get
\begin{align*}
\frac{s}{s-1}-s\int_1^N\frac{x-[x]}{x^{s+1}}\,dx &= \frac{s}{s-1}-s\int_1^Nx^{-s}\,dx+s\int_1^N\frac{[x]}{x^{s+1}}\,dx \\
&= \frac{s}{s-1}+\frac{s}{s-1}\biggl(\frac{1}{N^{s-1}}-1\biggr)+\Biggl(\sum_{n=1}^N\frac{1}{n^s}\Biggr)-\frac{1}{N^{s-1}} \\
&= \frac{1}{s-1}\biggl(\frac{1}{N^{s-1}}\biggr)+\Biggl(\sum_{n=1}^N\frac{1}{n^s}\Biggr)
\end{align*}Taking the limit as $N\rightarrow\infty$, we get the desired result.

Note that $$0<\int_1^b\frac{x-[x]}{x^{s+1}}\,dx<\int_1^b\frac{1}{x^{s+1}}\,dx,$$ and the integral on the right converges as $b\rightarrow\infty$ for all $s>0$ by the integral test of Exercise 8. Hence the integral in part (b) also converges as $b\rightarrow\infty$ for all $s>0$.


Exercise 17

(By analambanomenos) Following the hint, let $P=\{x_0,x_1,\ldots,x_n\}$ be a partition of $[a,b]$. We may assume that $g$ is real-valued. By the mean-value theorem, Theorem 5.10, each interval $(x_{i-1},x_i)$ contains $t_i$ such that $G(x_i)-G(x_i)=g(t_i)\Delta x_i$. Hence we have
\begin{align*}
& \sum_{i=1}^n\alpha(x_i)g(t_i)\Delta x_i \\
&\quad =\alpha(x_1)G(x_1)-\alpha(x_1)G(x_0)+\alpha(x_2)G(x_2)-\alpha(x_2)G(x_1)+\cdots+\alpha(x_n)G(x_n)-\alpha(x_n)G(x_{n-1}) \\
& \quad =-G(x_0)\alpha(x_0)-G(x_0)\bigl(\alpha(x_1)-\alpha(x_0)\bigr)-\cdots-G(x_{n-1})\bigl(\alpha(x_n)-\alpha(x_{n-1})\bigr)+G(x_n)\alpha(x_n) \\
& \quad =G(b)\alpha(b)-G(a)\alpha(a)-\sum_{i=1}G(x_{i-1})\Delta\alpha_i.
\end{align*}By Theorem 6.10, $\alpha g\in\mathscr R$, and by Theorem 6.8, $G\in\mathscr R(\alpha)$. Hence as the partition becomes finer, the sum on the right tends to $\int\alpha g\,dx$, and the sum on the left tends to $\int G\,d\alpha$.


Exercise 18

(By analambanomenos) Since for any $x\in\mathbf R$ we have $$|e^{ix}|=|\cos x+i\sin x|=\cos^2x+\sin^2x=1,$$ the images of all three curves lie on the unit circle $S^1$. And if $z=\cos x+i\sin x$ is an arbitrary point of $S^1$, $0\le x<2\pi$, then $z=\gamma_1(x)=\gamma_2(x/2)$, so the images of $\gamma_1$ and $\gamma_2$ are all of $S^1$. Also, since $$2\pi(\pi)\sin
\biggl(\frac{1}{\pi}\biggr)\approx 6.18\quad\quad 2\pi\biggl(\frac{2}{3\pi}\biggr)\sin\biggl(\frac{3\pi}{2}\biggr)\approx-1.33,$$ by the intermediate-value theorem (Theorem 4.23) if $-1.33\le y\le6.18$, there is a $0\le t\le 2\pi$ such that $2\pi t\sin(1/t)=y$, so that $\gamma_3(t)=e^{iy}=z$. Hence the image of $\gamma_3$ is also all of $S^1$.

Since $\big|\gamma_1(t)\big|=|ie^{it}|=1$ and $\big|\gamma_2(t)\big|=|2ie^{2it}|=2$, by Theorem 6.27 we have $$\Lambda(\gamma_1)=\int_0^{2\pi}1\,dt=2\pi\quad\quad\Lambda(\gamma_2)=
\int_0^{2\pi}2\,dt=4\pi.$$For $\gamma_3$ we have
\begin{align*}
\gamma_3′(t) &= 2\pi i\bigl(\sin(t^{-1})-t^{-1}\cos(t^{-1})\bigr)\gamma_3(t) \\
\big|\gamma_3′(t)\big| &= 2\pi\big|\sin(t^{-1})-t^{-1}\cos(t^{-1})\big|
\end{align*}Let $a_n=(2n+1)\pi$, $b_n=(2n+1/2)\pi$ for any integer $n\ge1$. Then $[a_n^{-1},b_n^{-1}]$ is a subinterval of $[0,2\pi]$ on which $\sin(1/t)\ge0$ and $\cos(1/t)\le0$. Hence the length of $\gamma_3$ on this subinterval is
\begin{align*}
\int_{a_n^{-1}}^{b_n^{-1}}\big|\gamma_3′(t)\big|\,dt &= 2\pi\int_{a_n^{-1}}^{b_n^{-1}}\sin(t^{-1})-t^{-1}\cos(t^{-1})\,dt \\
&= 2\pi\int_{a_n}^{b_n}-u^{-2}\sin(u)+u^{-1}\cos(u)\,du \\
&= 2\pi\bigl(b_n^{-1}\sin(b_n)-a_n^{-1}\sin(a_n)\bigr) \\
&= \frac{1}{n+(1/4)}
\end{align*} Since $\Lambda(\gamma_3)$ must be larger than the divergent sum of such terms, $\gamma_3$ is not rectifiable.


Exercise 19

(By analambanomenos) Suppose $\gamma_1$ is an arc. Since the composition of one-to-one mappings is clearly one-to-one, $\gamma_2$ is also an arc.

Note that $\phi$ is strictly monotonically increasing. For if not, there are points $c<x_1 < x_2 < d$ such that $\phi(c)=a < \phi(x_2) < \phi(x_1)$. by the intermediate-value theorem (Theorem 4.23) there is a point $x_3$, $c<x_3<x_1$ such that $\phi(x_3)=\phi(x_2)$, contradicting the fact that $\phi$ is one-to-one. Hence $\phi(d)=b$, so if $\gamma_1$ is a closed curve, then $$\gamma_2(d)=\gamma_1\bigl(\phi(d)\bigr)=\gamma_1(b)=\gamma_1(a)=\gamma_1\bigl(\phi(c)\bigr)=\gamma_2(c),$$ so that $\gamma_2$ is also closed.

Suppose that $\gamma_1$ is rectifiable. Let $P=\{x_0=a < x_2 <\cdots<x_n=c\}$ be a partition of $[a,b]$. Then since $\phi$ is strictly monotonically increasing, $\phi^{-1}(P)$ is a partition of $[c,d]$, and we have
\begin{align*}
\Lambda\bigl(\phi^{-1}(P),\gamma_2\bigr) &= \sum_{i=1}^n\Big|\gamma_2\bigl(\phi^{-1}(x_i)\bigr)-\gamma_2\bigl(\phi^{-1}(x_{i-1})\bigr)\Big| \\
&= \sum_{i=1}^n\big|\gamma_1(x_i)-\gamma_1(x_{i-1})\big| \\
&= \Lambda(P,\gamma_1).
\end{align*}Hence $\gamma_2$ is rectifiable if and only if $\gamma_1$ is rectifiable, in which case they have the same length.

Baby Rudin 数学分析原理完整第六章习题解答

Linearity

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