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Solution to Principles of Mathematical Analysis Chapter 10

Chapter 10 Integration of Differential Forms

Exercise 2

(By analambanomenos) The $\varphi_i(x)\varphi_i(y)$ has support in the square $2^{-i}<x<2{-1+1}$, $2^{-i}<y<2{-1+1}$, and the $\varphi_{i+1}(x)\varphi_i(y)$ term has support in the rectangle $2^{-i-1}<x<2{-1}$, $2^{-i}<y<2{-1+1}$, so $f$ has compact support in the square $0<x<1$, $0<y<1$. Each $(x,y)\ne(0,0)$ has a neighborhood small enough so that at most three of the terms in the sum are nonzero. Since these terms are continuous, $f$ is continous away from the origin.

Let $M_i$ be the maximum value of $\varphi_i$, attained at $x_i\in(2^{-i},2^{-i+1})$. Since $1=\int\varphi_i<M_i2^{-i}$, we have $M_i>2^i$. Hence $f(x_i,x_i)=M_i^2>2^{i+1}$ diverges to $\infty$ as $i\rightarrow\infty$, so $f$ is not continuous at $(0,0)$ and is unbounded in every neighborhood of $(0,0)$.

We have
\int dy\int f(x,y)\,dx &= \sum_{i=1}^\infty\Bigg(\int\varphi_i(y)\,dy\Bigg)\Bigg(\int\varphi_i(x)\,dx-\int\varphi_{i+1}(x)\,dx\Bigg) \\
&= \sum_{i=1}^\infty 1\cdot 0 = 0 \\
\int dx\int f(x,y)\,dy &= \Bigg(\int\varphi_1(x)\,dx\Bigg)\Bigg(\int\varphi_1(y)\,dy\Bigg)\;+ \\
&\qquad\sum_{i=2}^\infty\Bigg(\int\varphi_i(x)\,dx\Bigg)\Bigg(\int\varphi_i(y)\,dy-\int\varphi_{i-1}(y)\,dy\Bigg) \\
&= 1\cdot 1 + \sum_{i=2}^\infty 1\cdot 0 = 1

Exercise 3

(By analambanomenos)

(a) (Much of this solution just repeats the proof of Theorem 10.7.) Assume $1\le m\le n-1$, and make the following induction hypothesis (which evidently holds for $m=1$):

$V_m$ is a neighborhood of $\mathbf 0$, $\mathbf F_m\in\mathscr C’(V_m)$, $\mathbf F_m(\mathbf 0)=\mathbf 0$, $\mathbf F_m’(\mathbf 0)=I$, and for $\mathbf x\in V_m$,
P_{m-1}\mathbf F_m(\mathbf x)=P_{m-1}\mathbf x.
\end{equation}By \eqref{10.3.1}, we have
\mathbf F_m(\mathbf x)=P_{m-1}\mathbf x+\sum_{i=m}^n\alpha_i(\mathbf x)\mathbf e_i,
\]where $\alpha_m,\ldots,\alpha_n$ are real $\mathscr C’$-functions in $V_m$. Hence, for $j=m,\ldots,n$,
\mathbf e_j = \mathbf F_m’(\mathbf 0)\mathbf e_j = \sum_{i=m}^n(D_j\alpha_i)(\mathbf 0)\mathbf e_i.
\]Since $\mathbf e_m,\ldots\mathbf e_n$ are independent, we must have
(D_m\alpha_m)(\mathbf 0)=1\qquad(D_{m+1}\alpha_m)(\mathbf 0)=\cdots=(D_n\alpha_m)(\mathbf 0)=0.
\end{equation}Define, for $\mathbf x\in V_m$,
\mathbf G_m(\mathbf x)=\mathbf x+\big(\alpha_m(\mathbf x)-x_m\big)\mathbf e_m
\]Then $\mathbf G_m\in\mathscr C’(V_m)$, $\mathbf G_m$ is primitive, and $\mathbf G_m’(\mathbf 0)=I$ by \eqref{10.3.2}. The inverse function theorem shows therefore that there is an open set $U_m$, with $\mathbf 0\in U_m\subset V_m$, such that $\mathbf G_m$ is a 1-1 mapping of $U_m$ onto a neighborhood $V_{m+1}$ of $\mathbf 0$, in which $\mathbf G_m^{-1}$ is continuously differentiable, and
{\mathbf G_m^{-1}}’(\mathbf 0)=\mathbf G_m’(\mathbf 0)^{-1}=I.
\]Define $\mathbf F_{m+1}(\mathbf y)$, for $\mathbf y\in V_{m+1}$, by
\mathbf F_{m+1}(\mathbf y)=\mathbf F_m\circ\mathbf G_m^{-1}(\mathbf y).
\]Then $\mathbf F_{m+1}\in\mathscr C’(V_{m+1})$, $\mathbf F_{m+1}(\mathbf 0)=\mathbf 0$, and $\mathbf F_{m+1}’(\mathbf 0)=I$ by the chain rule. Also, for $\mathbf x\in U_m$,
P_m\mathbf F_{m+1}\big(\mathbf G_m(\mathbf x)\big) &= P_m\mathbf F_m(\mathbf x) \\
&= P_m\big(P_{m-1}\mathbf x+\alpha_m(\mathbf x)\mathbf e_m+\cdots\big) \\
&= P_{m-1}\mathbf x+\alpha_m(\mathbf x)\mathbf e_m \\
&= P_m\mathbf G_m(\mathbf x)
\end{align*}so that, for $\mathbf y\in V_{m+1}$, $P_m\mathbf F_{m+1}(\mathbf y)=P_m\mathbf y$. Our induction hypothesis holds therefore with $m+1$ in place of $m$.

Note that, for $\mathbf y=\mathbf G_m(\mathbf x)$, we have
\mathbf F_{m+1}\big(\mathbf G_m(\mathbf x)\big)=\mathbf F_m(\mathbf x).
\]If we apply this with $m=1,\ldots,n-1$, we successively obtain
\mathbf F_1 = \mathbf F_2\circ\mathbf G_1 = \mathbf F_3\circ\mathbf G_2\circ\mathbf G_1 = \cdots = \mathbf F_n\circ\mathbf G_{n-1}\circ\cdots\circ\mathbf G_1
\]in some neighborhood of $\mathbf 0$. By \eqref{10.3.1}, $\mathbf F_n$ is primitive, so we can let $\mathbf G_n=\mathbf F_n$.

(b) Let $\mathbf F$ be the mapping $(x,y)\rightarrow(y,x)$ and suppose $\mathbf F=\mathbf G_2\circ\mathbf G_1$ in some neighborhood of the origin, where
\mathbf G_1(x,y)=\big(f(x,y),y\big)\qquad\mathbf G_2(u,v)=\big(u,g(u,v)\big)
\]are primitive mappings. Then we would have
(y,x) &= \mathbf G_2\circ\mathbf G_1(x,y) \\
&= \mathbf G_2\big(f(x,y),y\big) \\
&= \big(f(x,y),g\big(f(x,y),y\big)\big)
\end{align*}so that
y=f(x,y)\qquad x=g\big(f(x,y),y\big)=g(y,y)
\]which is impossible. Trying $\mathbf F=\mathbf G_1\circ\mathbf G_2$ leads to a similar contradiction.

Exercise 4

(By analambanomenos) We have
\mathbf G_2\circ\mathbf G_1(x,y) &= \mathbf G_2(e^x\cos y-1,y) \\ &= (e^x\cos y-1,e^x\cos y\tan y) \\ &= (e^x\cos y-1,e^x\sin y) \\ &= \mathbf F(x,y) \end{align*}The derivative matrices are
\mathbf G_1′(x,y) &=
e^x\cos y & -e^x\sin y \\
0 & 1
\end{pmatrix} \\
\mathbf G_2′(u,v) &=
1 & 0 \\
\tan v & (1+u)\cos^{-2}v
\end{align*}so that $\mathbf G_1′(0,0)=\mathbf G_2′(0,0)=I$, hence $J_{\mathbf G_1}(0,0)=J_{\mathbf G_2}(0,0)=1$. By the chain rule and the properties of determinants, we also have $J_{\mathbf F}(0,0)=1$.

Let $h(u,v)=\sqrt{v^2-e^{2u}}-1$. Then, for $(x,y)$ near the origin,
\mathbf H_1\circ\mathbf H_2(x,y) &= \mathbf H_1(x,e^x\sin y) \\
&= \bigg(\sqrt{e^{2x}\sin^2y-e^{2x}}-1,e^x\sin y\bigg) \\
&=(e^x\cos y-1,e^x\sin y) \\
&= \mathbf F(x,y)

Exercise 5

(By analambanomenos) We want to show: Suppose $K$ is a compact subset of a metric space $X$, and $\{V_\alpha\}$ is an open cover of $K$. Then there exists $\psi_1,\ldots,\psi_s\in\mathscr C(X)$ such that

(a) $0\le\psi_i\le 1$ for $1\le i\le s$;
(b) each $\psi_i$ has its support in some $V_\alpha$, and
(c) $\psi_i(x)+\cdots+\psi_s(x)=1$ for every $x\in K$.

Repeating the proof of Theorem 10.8 in the text and following the hint, associate with each $x\in K$ an index $\alpha(x)$ so that $x\in V_{\alpha(x)}$. Then there are open balls $B(x)$ and $W(x)$ centered at $x$, with
\overline{B(x)}\subset W(x)\subset\overline{W(x)}\subset V_{\alpha(x)}.
\]Since $K$ is compact, there are points $x_1,\ldots,x_s$ in $K$ such that
K\subset B(x_1)\cup\cdots\cup B(x_s).
\]By Exercise 4.22, there are functions $\varphi_1\,\ldots,\varphi_s\in\mathscr C(X)$ such that $\varphi_i(x)=1$ on $\overline{B(x_i)}$, $\varphi_i(x)=0$ outside $W(x_i)$, and $0\le\varphi_i(x)\le 1$ on $X$, namely,
\]where $\rho_{i1}(x)$ is the distance from $x$ to the complement of $W(x_i)$, a closed set, and $\rho_{i2}(x)$ is the distance from $x$ to $\overline{B(x_i)}$. Letting $\psi_1=\varphi_1$, and
\]for $i=1,\ldots,s-1$, the remainder of the proof follows exactly as in the proof of Theorem 10.8.

Exercise 6

(By analambanomenos) Following the hint, recall that Exercise 8.1 defined an infinitely differentiable function on $\mathbf R^1$ such that $f(x)=0$ for $x\le0$ and $0\le f(x)<1$ for all $x$. Let $a<b$. Then the function $g_{a,b}(x)=f(x-a)f(b-x)$ is also infinitely differentiable, equals 0 for $x\le a$ and $x\ge b$, and $0\le g_{a,b}(x)<1$ for all $x$. Since it has compact support, we can define a function
h_{a,b}(x)=\frac{1}{A}\int_x^\infty g_{a,b}(t)\,dt\qquad\hbox{where}\qquad A=\int_{-\infty}^\infty g_{a,b}(t)\,dt
\]which is infinitely differentiable, equals 1 for $x\le a$, equals 0 for $x\ge b$, and $0\le h_{a,b}(x)\le 1$ for all $x$.

Now let $\mathbf x\in\mathbf R^n$, and let $B(\mathbf x)$ and $W(\mathbf x)$ be open balls centered at $\mathbf x$ with radii $a<b$, respectively. Define the function $r(\mathbf y)$ for $\mathbf y\in\R n$ which is the distance between $\mathbf x$ and $\mathbf y$, that is,
r(\mathbf y)=\sqrt{\sum(x_i-y_i)^2}
\]which is infinitely differentiable for $\mathbf y\ne\mathbf x$. Then the function $\varphi=h_{a,b}\circ r$ is infinitely differentiable on $\R n$, equals 1 on $\overline{B(\mathbf x)}$, equals 0 on $W(\mathbf x)$, and $0\le\varphi(\mathbf y)\le 1$ for all $\mathbf y$. We can use these functions in the proof of Theorem 10.8 to get infinitely differentiable functions $\psi_i$.

Exercise 7

(By analambanomenos)

(a) First we need to show that $Q^k$ is convex. Let $\mathbf x,\mathbf y\in Q^k$, so that the components satisfy
x_i\ge 0,\quad y_i\ge 0,\quad\sum x_i\le 1,\quad\sum y_i\le 1.
\]Let $0\le\lambda\le 1$, and let $\mathbf z=\lambda\mathbf x+(1-\lambda)\mathbf y$. Then
z_i=\lambda x_i+(1-\lambda)y_i\qquad\sum z_i=\lambda\sum x_i+(1-\lambda)\sum y_i
\]so that $z_i$ lies between $x_i$ and $y_i$, and $\sum z_i$ lies between $\sum x_i$ and $\sum y_i$. Hence $\mathbf z\in Q^k$.

Let $C$ be a convex subset of $\mathbf R^k$ containing $\mathbf 0,\mathbf e_1,\ldots,\mathbf e_k$; we need to show that $Q^k\subset C$. We can consider $Q^i\subset Q^j$ for $i<j$ by letting the components with index greater than $i$ be 0. I am going to show that $Q^i\subset C$, $i=1,\ldots,k$ by induction. Let $\mathbf x\in Q^1$. Then $\mathbf x=x_1\mathbf e_1+(1-x_1)\mathbf 0$ for $0\le x_1\le 1$, so that $\mathbf x\in C$. Now suppose that $Q^{i-1}\subset C$ and let $\mathbf x\in Q^i$. Then $x_1+\ldots+x_i\le 1$ implies
\frac{(x_1+\ldots+x_{i-1})}{1-x_i}\le 1.
\]so that
\mathbf x’=(1-x_i)^{-1}(x_1,\ldots,x_{i-1},0,\ldots,0)\in Q^{i-1}\subset C.
\]Hence $\mathbf x=(1-x_i)\mathbf x’+x_i\mathbf e_i\in C$ since $0\le x_i\le 1$, which shows that $Q^i\subset C$.

(b) Let $X,Y$ be vector spaces and let $\mathbf f=\mathbf f(\mathbf 0)+A$, for some $A\in L(X,Y)$, be an affine mapping from $X$ to $Y$. Let $C$ be a convex subset of $X$, and let $\mathbf y_1=\mathbf f(\mathbf x_1)$, $\mathbf y_2=\mathbf f(\mathbf x_2)$ be elements of $\mathbf f(C)$ for some $\mathbf x_1\in C$ and $\mathbf x_2\in C$. Then for $0\le\lambda\le 1$, we have $$\lambda\mathbf x_1+(1-\lambda)\mathbf x_2\in C,$$ so that
\lambda\mathbf y_1+(1-\lambda)\mathbf y_2 = \mathbf f(\mathbf 0)+\lambda A(\mathbf x_1)+(1-\lambda)A(\mathbf x_2) = \mathbf f(\mathbf 0)+A\big(\lambda\mathbf x_1+(1-\lambda)\mathbf x_2\big)\in\mathbf f(C).
\]Hence $\mathbf f(C)$ is convex.

Exercise 8

(By analambanomenos) Since $(3,2)=(1,1)+(2,1)$ and $(2,4)=(1,1)+(1,3)$, the linear part of the affine map is $A(u,v)=(2u+v,u+3v)$, so
T(u,v) &= (1,1)+A(u,v)=(2u+v+1,u+3v+1) \\
J_T &=
2 & 1 \\
1 & 3
=5 \\
\int_He^{x-y}\,dx\,dy &= \int_0^1\int_0^1e^{(2u+v+1)-(u+3v+1)}J_T\,du\,dv \\
&= 5\bigg(\int_0^1e^u\,du\bigg)\bigg(\int_0^1e^{-2v}\,dv\bigg) \\
&= \textstyle\frac{5}{2}(e-e^{-1}+e^{-2}-1)

Baby Rudin 数学分析原理不完整第十章习题解答


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