If you find any mistakes, please make a comment! Thank you.

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
||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
||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

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
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
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
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
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
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
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
\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
& \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
\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
\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
\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


This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has One Comment

  1. thanks for sharing. I find this better than Cook's solutions, specially because I´'m working on all the problems by myself. I have problems with my eyes. Do you think it would be possible to get a pdf version please so that i can print it out? thanks in anycase.

Leave a Reply

Close Menu