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Analysis on Manifolds 11. Existence of the Integral

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本节讨论了积分存在的一个充要条件，即Theorem 11.2，不连续点的测度为0是积分存在的充要条件，为此Theorem 11.1首先定义了零测度，并给出了一些有用的关于零测度的结论。Theorem 11.3是主要定理的一个直接应用，讨论f vanished except on a set of measure zero 和\int_Qf之间的关系。

Exercises

Exercise 1. Show that if A has measure zero in \mathbf{R}^n, the sets \overline{A} and \text{Bd }A need not have measure zero

Solution: Let A={(q_1,q_2,\dots,q_n):q_i\in Q\cap [0,1]}, then A has measure zero in \mathbf{R}^n, while \overline{A}=\text{Bd }A=[0,1]\times\cdots\times[0,1], which is a rectangle in \mathbf{R}^n.

Exercise 2. Show that no open set in \mathbf{R}^n has measure zero in \mathbf{R}^n.

Solution: For A in \mathbf{R}^n which is open, choose \mathbf{a}\in A, then there is arectangle Q which centered in \mathbf{a} such that Q\subset A, since 0<v(Q)\leq v(A) the conclusion follows.

Exercise 3. Show that the set \mathbf{R}^{n-1}\times 0 has measure zero in \mathbf{R}^n.

Solution: It suffice to show the set A=[0,1]\times\cdots\times[0,1]\times 0 has measure zero, and the conclusion follows from Theorem 11.1(b). Given \epsilon>0, let Q=[0,1]\times\cdots\times[0,1]\times[-\epsilon/4,\epsilon/4], then Q covers the A and v(Q)=\epsilon/2<\epsilon.

Exercise 4. Show that the set of irrationals in [0,1] does not have measure zero in \mathbf{R}.

Solution: We already know the set of rationals in [0,1] has measure zero in \mathbf{R}, assume the set of irrationals in [0,1] does not have measure zero in \mathbf{R}, then the two sets combined covers [0,1], use Theorem 11.1( c ) we can find a countable covering of both the set of rationals in [0,1] and the set of irrationals in [0,1] by open rectangles \text{Int }Q_1,\text{Int }Q_2,\dots such that \sum_{i=1}^{\infty}v(Q_i)<1/2, this covering of sets also cover [0,1] which is compact, thus we can find a finite cover, whose total volume when summing is less than 1/2, but v([0,1])=1.

Exercise 5. Show that if A is a compact subset of \mathbf{R}^n and A has measure zero in \mathbf{R}^n, then given \epsilon>0, there is a finite collection of rectangles of total volume less than \epsilon covering A.

Solution: Use Theorem 11.1( c ), we first find a countable covering of A by \text{Int }Q_1,\text{Int }Q_2,\dots such that \sum_{i=1}^{\infty}v(Q_i)<\epsilon, as A is compact, we can find a finite cover \text{Int }Q_{i_1},\dots\text{Int }Q_{i_n} which covers A, and \sum_{j=1}^{n}v(Q_{i_j})<\epsilon, obviously Q_{i_1},\dots Q_{i_n} covers A.

Exercise 6. Let f:[a,b]\to\mathbf{R}. The graph of f is the subset

\displaystyle{G_f=\{(x,y)|y=f(x)\}}

of \mathbf{R}^2. Show that if f is continuous, G_f has measure zero in \mathbf{R}^2.

Solution: If f is continuous, then f is integrable over [a,b], thus given \epsilon>0, there is a partition P such that U(f,P)-L(f,P)<\epsilon, denote P=(x_0,x_1,\dots,x_n,x_{n+1}), in which x_0=a,x_{n+1}=b, let m_i be the point in [x_i,x_{i+1}] in which f(x)\geq f(m_i) for x\in [x_i,x_{i+1}], and M_i be the point in [x_i,x_{i+1}] in which f(x)\leq f(M_i) for x\in [x_i,x_{i+1}], let Q_i=[x_i,x_{i+1}]\times[m_i,M_i], then the rectangles Q_1,\dots,Q_n covers G_f, and \sum_i v(Q_i)=U(f,P)-L(f,P).

Exercise 7. Consider the function f defined in Example 2. At what points of [0,1] does f fail to be continuous? Answer the same question for the funtion defined in Exercise 5 of Section 10.

Solution: For Example 2, f fails to be continuous at all points in [0,1]. For Exercise 5 of Section 10, f fails to be continuous at all rational points in [0,1].

Exercise 8. Let Q be a rectangle in \mathbf{R}^n; let f:Q\to\mathbf{R} be a bounded function. Show that if f vanishes except on a closed set B of measure zero, then \int_Qf exists and equals zero.
Solution: We suppose |f|\leq M on Q, for any \epsilon >0, we choose a countable collection Q_1,Q_2,\dots which covers B and \sum_i v(Q_i)<\epsilon/2M , since B is bounded and closed, B is compact and we can choose finite Q_is to cover B, we denote those finite Q_is as Q_1',\dots,Q_n', these together with Q-B, form a partition P of Q, we can see that

\displaystyle{U(f,P)-L(f,P)\leq 2M \sum_{i=1}^nv(Q_i')<2M\sum_iv(Q_i)=\epsilon}

It follows that \int_Qf exists, and the conclusion follows from Theorem 11.3.

Exercise 9. Let Q be a rectangle in \mathbf{R}^n; let f:Q\to\mathbf{R}; assume f is integrable over Q.
( a ) Show that if f(\mathbf{x})\geq 0 for \mathbf{x}\in Q, then \int_Qf\geq 0.
( b ) Show that if f(\mathbf{x})> 0 for \mathbf{x}\in Q, then \int_Qf>0.

Solution:
( a ) L(f,P)\geq 0 for any partition P, thus \int_Qf\geq 0.
( b ) If we assume \int_Qf=0, then by Theorem 11.3(b), f vanishes except on a set of measure zero, but Q does not have measure zero.

Exercise 10. Show that if Q_1,Q_2,\dots is a countable collection of rectangles covering Q, then v(Q)\leq\sum v(Q_i).

Solution: We have

\displaystyle{v(Q)=v(Q\cap\bigcup_iQ_i)=v\left(\bigcup_i(Q\cap Q_i)\right)\leq\sum_iv(Q\cap Q_i)\leq\sum_iv(Q_i)}

Analysis on Manifolds 10. The Integral Over a Rectangle

christangdt留下评论

本书对多元积分的处理是先推广一元黎曼积分，本节仅对黎曼积分的一些大家熟知的概念作了推广，包括partition以及黎曼和等。比较重要的是Theorem 10.3(Riemann condition)，下积分和上积分相等的充要条件，与一元黎曼积分几乎完全类似。Theorem 10.4是黎曼条件的一个小应用，证明常数函数是可积的。

Exercises

Exercise 1. Let f,g:Q\to\mathbf{R} be bounded functions such that f(\mathbf{x})\leq g(\mathbf{x}) for \mathbf{x}\in Q. Show that \underline{\int}_Qf\leq\underline{\int}_Qg and \overline{\int}_Qf\leq\overline{\int}_Qg.

Solution: As \underline{\int}_Qf=\sup_P{L(f,P)}, for any \epsilon>0, we can find a partition P such that

\displaystyle{\underline{\int}_Qf-\epsilon U(g,P')\geq U(f,P')\geq \overline{\int}_Qf}

and it follows that \overline{\int}_Qf\leq\overline{\int}_Qg.

Exercise 2. Suppose f:Q\to\mathbf{R} is continuous. Show f is integrable over Q.

Solution: We suppose Q\in\mathbf{R}^n, then Q is bounded, and f is in fact uniformly continuous on Q, thus for any \epsilon>0, there is a \delta>0 such that

\displaystyle{|\mathbf{x}-\mathbf{y}|<\delta\implies |f(\mathbf{x})-f(\mathbf{y})|<\frac{\epsilon}{v(Q)}}

Choose a partition P with \text{mesh }P_i<\delta/\sqrt{n} for i=1,\dots,n, then within each subrectangle contained in P, f would have diversion at most \epsilon/v(Q), thus

\displaystyle{U(f,P)-L(f,P)<\frac{\epsilon}{v(Q)}\sum_{R\subset P}v(R)=\epsilon}

Exercise 3. Let [0,1]^2=[0,1]\times [0,1]. Let f:[0,1]^2\to\mathbf{R} be defined by setting f(x,y)=0 if x\neq y, and f(x,y)=1 if y=x. Show that f is integrable over [0,1]^2.

Solution: For any \epsilon>0, let N=[1/\epsilon]+1, and P'={0,1/N,\dots,(N-1)/N,1}, let P=(P',P'), we see that U(f,P) and L(f,P) only diverges in rectangles R_k=[k/N,(k+1)/N]\times [k/N,(k+1)/N], in which k ranges from 0 to N-1 and v(R_k)=1/N^2, thus

\displaystyle{U(f,P)-L(f,P)=N\frac{1}{N^2}=\frac{1}{N}<\epsilon}

Exercise 4. We say f:[0,1]\to\mathbf{R} is increasing if f(x_1)\leq f(x_2) whenever x_1<x_2. If f,g:[0,1]\to\mathbf{R} are increasing and non-negative, show that the function h(x,y)=f(x)g(y) is integrable over [0,1]^2.

Solution: Both f and g are integrable over [0,1], which means \int_{[0,1]}f and \int_{[0,1]}g exist.
For any \epsilon>0, let n be an integer and P'={0,1/n,\dots,(n-1)/n,1}, let P=(P',P'), we see that

\displaystyle{U(f,P)-L(f,P)=\frac{1}{n^2}\left[(f(1)-f(0))\left(g\left(\frac{1}{n}\right)+\cdots+g(1)\right)+(g(1)-g(0))\left(f\left(\frac{1}{n}\right)+\cdots+f(1)\right)\right]}

Notice that we can choose N such that when n>N,

\displaystyle{\left|\frac{g(1/n)+\cdots+g(1)}{n}-\int_{[0,1]}g\right|<1,\left|\frac{f(1/n)+\cdots+f(1)}{n}-\int_{[0,1]}f\right|<1}

thus the expression

\displaystyle{\frac{1}{n}\left[(f(1)-f(0))\left(g\left(\frac{1}{n}\right)+\cdots+g(1)\right)+(g(1)-g(0))\left(f\left(\frac{1}{n}\right)+\cdots+f(1)\right)\right]}

is in fact bounded, we can choose n large enough to guarantee U(f,P)-L(f,P)<\epsilon, and the conclusion follows.

Exercise 5. Let f:\mathbf{R}\to\mathbf{R} be defined by setting f(x)=1/q if x=p/q, where p and q are positive integers with no common factor, and f(x)=0 otherwise, show f is integrable over [0,1].

Solution: Since f only take nonzero values on Q\cap[0,1], we see that L(f,P)=0 for any partition P of [0,1]. For any \epsilon>0, let n>1/\epsilon and P'=(0,1/n,\dots,1) be a partition of [0,1], notice that there will be only n+1 rectangles in the sum of U(f,P), and the value 1/q would be the supremum of f for some rectangle only if all the value 1,1/2,\dots,1/(q-1) have been the supremum of f for some rectangle in P, and each value 1/k would occur k-1 times for k>1, thus each rectangle which take the supremum value 1/q would amount at most q-1, and the total contribution in the sum U(f,P) would be \frac{q-1}{q}\frac{1}{n}, thus U(f,P)<1/n.

Exercise 6. Prove the following:
Theorem. Let f:Q\to\mathbf{R} be bounded. Then f is integrable over Q if and only if given \epsilon>0, there is a \delta>0 such that U(f,P)-L(f,P)<\epsilon for every partition P of mesh less than \delta.
Proof. ( a ) Verify the “if” part of the theorem.
( b ) Suppose |f(\mathbf{x})|\leq M for \mathbf{x}\in Q. Let P be a partition of Q. Show that if P'' is obtained by adjoining a single point to the partition of one of the component intervals of Q, then

\displaystyle{0\leq L(f,P'')-L(f,P)\leq 2M(\text{mesh }P)(\text{width }Q)^{n-1}.}

Derive a similar result for upper sums.
( c ) Prove the “only if” part of the theorem: Suppose f is integrable over Q. Given \epsilon>0, choose a partition P' such that U(f,P')-L(f,P')<\epsilon/2. Let N be the number of partition points in P'; then let \delta=\epsilon/8MN(\text{width }Q)^{n-1}. Show that if P has mesh less than \delta, then U(f,P)-L(F,P)<\epsilon.

Solution:
( a ) The “if” part is easy from the definition.
( b ) If P'' is obtained by adjoining a single point to the partition of one of the component intervals [a_i,b_i] of Q, if P=(P_i,\dots,P_n), without loss of generality we suppose this single point is added in the interval I\in P_1, thenL(f,P'') and L(f,P) only differs in subrectangles with the form
\displaystyle{R=I\times I_2\times\cdots\times I_n\in P, I_i\in P_i \text{ for } i=2,\dots,n}
The point divides I into I' and I'', let L(I) denote the length of an interval I, we have L(I)=L(I')+L(I''). For an specific subrectangle R which is being divided, we see that

\displaystyle{R=R'+R'',R'=I'\times I_2\times\cdots\times I_n,R''=I''\times I_2\times\cdots\times I_n}

If f reaches local minimum at a\in R and a\in R', then f reaches local minimum at a in R', vice versa, in the first case we suppose in R'' the local minimum of f occurs at a', then the contribution in L(f,P) takes the form

\displaystyle{V_1=f(a)L(I)L(I_2)\cdots L(I_n)=f(a)L(I)v( I_2\times\cdots\times I_n)}

while the contribution in L(f,P'') takes the form

\displaystyle{V_2=f(a)L(I')v( I_2\times\cdots\times I_n)+f(a')L(I'')v( I_2\times\cdots\times I_n)}

Compare the difference we have

\displaystyle{\begin{aligned}V_2-V_1&=f(a)[L(I')-L(I)]v( I_2\times\cdots\times I_n)+f(a')L(I'')v( I_2\times\cdots\times I_n)\\&=[f(a')-f(a)]L(I'')v( I_2\times\cdots\times I_n)\\&\leq2M(\text{mesh }P)v( I_2\times\cdots\times I_n)\end{aligned}}

As I_i goes through P_i, we have

\displaystyle{L(f,P'')-L(f,P)\leq 2M(\text{mesh }P)\sum_{I_i\in P_i}v( I_2\times\cdots\times I_n)\leq 2M(\text{mesh }P)(\text{width }Q)^{n-1}}

A similar result can be derived for upper sums, i.e. if P'' is obtained by adjoining a single point to the partition of one of the component intervals of Q, then

\displaystyle{0\leq U(f,P)-U(f,P'')\leq 2M(\text{mesh }P)(\text{width }Q)^{n-1}.}

( c ) If P has mesh less than \delta, then we add the N points of P' to get a refinement of P and P', namely P'', we have

\displaystyle{0\leq L(f,P'')-L(f,P)\leq 2MN(\text{mesh }P)(\text{width }Q)^{n-1}<\epsilon/4}

Similarly we have 0\leq U(f,P)-U(f,P'')<\epsilon/4, thus using the fact that P'' is a refinement of P‘, we have

\displaystyle{U(f,P)-L(f,P)<U(f,P')-L(f,P')+\epsilon/2<\epsilon}

Exercise 7. Use Exercise 6 to prove the following:
Theorem. Let f:Q\to\mathbf{R} be bounded. Then the statement that f is integrable over Q, with \int_Qf=A, is equivalent to the statement that given \epsilon>0, there is a \delta>0 such that if P is any partition of mesh less than \delta, and if for each subrectangle R determined by P, x_R is a point of R, then

\displaystyle{|\sum_Rf(\mathbf{x}_R)v(R)-A|<\epsilon}

Solution: If f is integrable over Q, then by Exercise 6, given \epsilon>0 there is a \delta>0 such that if P is any partition of mesh less than \delta, we have U(f,P)-L(f,P)<\epsilon, notice that L(f,P)\leq A\leq U(f,P) and for each subrectangle R determined by P we have

\displaystyle{L(f,P)\leq\sum_Rf(\mathbf{x}_R)v(R)\leq U(f,P)}

the “if” part of the theorem is proved. Conversely, given \epsilon>0, we first find a \delta>0 such that for each subrectangle R determined by P which has mesh less than \delta:

\displaystyle{|\sum_Rf(\mathbf{x}_R)v(R)-A|<\epsilon/2,\quad\mathbf{x}_R\in R}

then |U(f,P)-A|<\epsilon/2 and |L(f,P)-A|<\epsilon/2, which means U(f,P)-L(f,P)<\epsilon, so f is integrable by Exercise 6. Assume \int_Qf=B\neq A, then let \epsilon=|B-A|/2 we can easily deduce a contradiction.

Analysis on Manifolds 9. The Implicit Function Theorem

christangdt留下评论

本节是另一个重要的定理：隐函数定理，隐函数表示的函数在某一点(a,b)可微的充要条件是隐函数关于因变量的导数在(a,b)点是非奇异的，并且可以显式计算出隐函数的导数。

Exercises

Exercise 1. Let f:\mathbf{R}^3\to\mathbf{R}^2 be of class C^1; write f in the form f(x,y_1,y_2). Assume that f(3,-1,2)=0 and

\displaystyle{Df(3,-1,2)=\begin{bmatrix}1&2&1\\1&-1&1\end{bmatrix}.}

( a ) Show there is a function g:B\to\mathbf{R}^2 of class C^1 defined on an open set B in \mathbf{R} such that f(x,g_1(x),g_2(x))=0 for x\in B, and g(3)=(-1,2).
( b ) Find Dg(3).
( c ) Discuss the problem of solving the equation f(x,y_1,y_2)=0 for an arbitrary pair of the unknowns in terms of the third, near the point (3,-1,2).

Solution:
( a ) Since

\displaystyle{\det\frac{\partial f(x,y_1,y_2)}{\partial(y_1,y_2)}(3,-1,2)=\det\begin{bmatrix}2&1\\-1&1\end{bmatrix}=5\neq 0}

by the implicit function theorem, the conclusion holds.
( b ) We have

\displaystyle{\begin{aligned}Dg(3)&=-\left[\frac{\partial f(x,y_1,y_2)}{\partial(y_1,y_2)}(3,-1,2)\right]^{-1}\cdot\frac{\partial f(x,y_1,y_2)}{\partial(x)}(3,-1,2)\\&=-\begin{bmatrix}2&1\\-1&1\end{bmatrix}^{-1}\cdot\begin{bmatrix}1\\1\end{bmatrix}\\&=\frac{1}{3}\begin{bmatrix}-1&1\\-1&2\end{bmatrix}\cdot\begin{bmatrix}1\\1\end{bmatrix}=\frac{1}{3}\begin{bmatrix}0\\1\end{bmatrix}\end{aligned}}

( c ) By some easy calculation we can know that

\displaystyle{\det\frac{\partial f(x,y_1,y_2)}{\partial(y_1,y_2)}(3,-1,2)\neq 0,\quad\det\frac{\partial f(x,y_1,y_2)}{\partial(x,y_1)}(3,-1,2)\neq0}

thus the equation f(x,y_1,y_2)=0 can be solved for (y_1,y_2) in terms of x, and for (x,y_1) in terms of y_2, however, as

\displaystyle{\det\frac{\partial f(x,y_1,y_2)}{\partial(x,y_2)}(3,-1,2)=\det\begin{bmatrix}1&1\\1&1\end{bmatrix}=0}

the equation f(x,y_1,y_2)=0 cannot be solved for (x,y_2) in terms of y_1.

Exercise 2. Given f:\mathbf{R}^5\to\mathbf{R}^2, of class C^1. Let \mathbf{a}=(1,2,-1,3,0); suppose that f(\mathbf{a})=\mathbf{0} and

\displaystyle{Df(\mathbf{a})=\begin{bmatrix}1&3&1&-1&2\\0&0&1&2&-4\end{bmatrix}.}

( a ) Show there is a function g:B\to\mathbf{R}^2 of class C^1 defined on an open set B of \mathbf{R}^3 such that

\displaystyle{f(x_1,g_1(x),g_2(x),x_2,x_3)=0}

for x=(x_1,x_2,x_3)\in B, and g(1,3,0)=(2,-1).
( b ) Find Dg(1,3,0).
( c ) Discuss the problem of solving the equation f(\mathbf{x})=\mathbf{0} for an arbitrary pair of the unknowns in terms of the others, near the point \mathbf{a}.
Solution:
( a ) Since

\displaystyle{\det\frac{\partial f}{\partial(g_1,g_2)}(\mathbf{a})=\det\begin{bmatrix}3&1\\0&1\end{bmatrix}=3\neq 0}

by the implicit function theorem, the conclusion holds.
( b ) We have

\displaystyle{\begin{aligned}Dg(1,3,0)&=-\left[\frac{\partial f}{\partial(g_1,g_2)}(\mathbf{a})\right]^{-1}\cdot\frac{\partial f}{\partial(x_1,x_2,x_3)}(\mathbf{a})\\&=-\begin{bmatrix}3&1\\0&1\end{bmatrix}^{-1}\cdot\begin{bmatrix}1&-1&2\\0&2&-4\end{bmatrix}\\&=\frac{1}{3}\begin{bmatrix}-1&1\\0&-3\end{bmatrix}\cdot\begin{bmatrix}1&-1&2\\0&2&-4\end{bmatrix}\\&=\begin{bmatrix}-1/3&1&-2\\0&-2&4\end{bmatrix}\end{aligned}}

( c ) The pairs that cannot solve the equation f(\mathbf{x})=\mathbf{0} in terms of the others, denoting \mathbf{x}=(x_1,x_2,x_3,x_4,x_5), are: (x_1,x_2) and (x_4,x_5).

Exercise 3. Let f:\mathbf{R}^2\to\mathbf{R} be of class C^1, with f(2,-1)=-1. Set

\displaystyle{\begin{aligned}G(x,y,u)&=f(x,y)+u^2,\\H(x,y,u)&=ux+3y^3+u^3.\end{aligned}}

The equations G(x,y,u)=0 and H(x,y,u)=0 have the solution (x,y,u)=(2,-1,1).
( a ) What conditions on Df ensure that there are C^1 function x=g(y) and u=h(y) defined on an open set in \mathbf{R} that satisfy both equations, such that g(-1)=2 and h(-1)=1?
( b ) Under the conditions of (a), and assuming that Df(2,-1)=[1,-3], find g'(-1) and h'(-1).
Solution:
( a ) We have DG=\begin{bmatrix}\partial{f}/\partial{x}&\partial{f}/\partial{y}&2u\end{bmatrix} and DH=\begin{bmatrix}2u&9y^2&x+3u^2\end{bmatrix}, thus

\displaystyle{\frac{\partial{G}}{\partial{y}}(2,-1,1)=\frac{\partial f}{\partial y}(2,-1),\quad \frac{\partial{H}}{\partial{y}}(2,-1,1)=9}

thus we need \partial{f}/\partial{y}(2,-1)\neq 0.
( b ) From G(2,-1,1)=0 we have

\displaystyle{\frac{\partial{f}}{\partial{x}}(2,-1)g'(-1)+\frac{\partial{f}}{\partial{y}}(2,-1)+2h(-1)h'(-1)=0}

which gives g'(-1)+2h'(-1)=3. Similarly, from H(2,-1,1)=0 we have

\displaystyle{h(-1)g'(-1)+h'(-1)g(-1)+9(-1)^2+3[h(-1)]^2h'(-1)=0}

which gives g'(-1)+5h'(-1)=-9. Together we solve to get g'(-1)=15,h'(-1)=-6.

Exercise 4. Let F:\mathbf{R}^2\to\mathbf{R} be of class C^2, with F(0,0)=0 and DF(0,0)=[2,3], Let G:\mathbf{R}^3\to\mathbf{R} be defined by the equation

\displaystyle{G(x,y,z)=F(x+2y+3z-1,x^3+y^2-z^2)}

( a ) Note that G(-2,3,-1)=F(0,0)=0. Show that one can solve the equation G(x,y,z)=0 for z, say z=g(x,y), for (x,y) in a neighborhood B of (-2,3), such that g(-2,3)=-1.
( b ) Find Dg(-2,3).
( c ) If D_1D_1F=3 and D_1D_2F=-1 and D_2D_2F=5 at (0,0), find D_2D_1g(-2,3).
Solution:
( a ) Let H(x,y,z)=(x+2y+3z-1,x^3+y^2-z^2), then G(x,y,z)=(F\circ H)(x,y,z), and by the chain rule

\displaystyle{DG(-2,3,-1)=DF(0,0)\cdot DH(-2,3,-1)=[2,3]\cdot\begin{bmatrix}1&2&3\\12&6&2\end{bmatrix}=\begin{bmatrix}38&22&12\end{bmatrix}}

We can see that \partial G/\partial z(-2,3,1)=12, which satisfies the condition of the implicit function theorem.
( b ) Dg(-2,3)=-\dfrac{1}{12}[38,22].
( c ) We have DF=[D_1F,D_2F] and

\displaystyle{DH=\begin{bmatrix}1&2&3\\3x^2&2y&-2z\end{bmatrix}}

this gives

\displaystyle{DG(x,y,z)=DF\cdot DH=\begin{bmatrix}D_1F+3x^2D_2F&2D_1F+2yD_2F&3D_1F-2zD_2F\end{bmatrix}}

from this and the expression of Dg(x,y) we have, suppose the condition of implicit funtion theorem is satisfied, that for z=g(x,y):

\displaystyle{Dg(x,y)=-\frac{1}{3D_1F-2zD_2F}\begin{bmatrix}D_1F+3x^2D_2F&2D_1F+2yD_2F\end{bmatrix}}
\displaystyle{D_1g(x,y)=-\frac{D_1F+3x^2D_2F}{3D_1F-2zD_2F}}

thus to calculate D_2D_1g(-2,3), we notice D_1F(0,0)=2,D_2F(0,0)=3, and as F is of class C^2, we get D_1D_2F=D_2D_1F=-1 at (0,0). Let H=D_1F+3x^2D_2F, we have D_2H=D_2D_1F+3x^2D_2D_2F and

\displaystyle{H(-2,3,1)=2+36=38,D_2H(-2,3,1)=-1+60=59}

Let K=3D_1F-2zD_2F, we have D_2K=3D_2D_1F-2D_2gD_2F-2gD_2D_2F, thus

\displaystyle{K(-2,3,1)=6-2(-1)3=12,D_2K(-2,3,1)=3(-1)+11+2\times 5=18}

Now

\displaystyle{D_2D_1g(-2,3)=-\frac{(D_2H)K-H(D_2K)}{K^2}(-2,3,1)=-\frac{59\times 12-38\times 18}{144}=-\frac{24}{144}}

Exercise 5. Let f.g:\mathbf{R}^3\to\mathbf{R} be functions of class C^1. “In general”, one expects that each of the equations f(x,y,z)=0 and g(x,y,z)=0 represents a smooth surface in \mathbf{R}^3, and that their intersection is a smooth curve. Show that if (x_0,y_0,z_0) satisfies both equations, and if \partial (f,g)/\partial (x,y,z) has rank 2 at (x_0,y_0,z_0), then near (x_0,y_0,z_0), one can solve these equations for two of x,y,z in terms of the third, thus representing the solution set locally as a parametrized curve.
Solution: Since \partial (f,g)/\partial (x,y,z) has rank 2 at (x_0,y_0,z_0), we see that at least one of \det\partial (f,g)/\partial (x,z), \det\partial (f,g)/\partial (x,y) and \det\partial (f,g)/\partial (y,z) is nonzero, which means we can apply the implicit function theorem to get the result.

Exercise 6. Let f:\mathbf{R}^{k+n}\to\mathbf{R}^n be of class C^1; suppose that f(\mathbf{a})=\mathbf{0} and that Df(\mathbf{a}) has rank n. Show that if \mathbf{c} is a point of \mathbf{R}^n sufficiently close to \mathbf{0}, then the equation f(\mathbf{x})=\mathbf{c} has a solution.
Solution: Since Df(\mathbf{a}) has rank n, we can find n columns of Df(\mathbf{a}) to be linearly independent, we assume the first n columns are so, and discuss the general case at last.
Let F:\mathbf{R}^{k+2n}\to\mathbf{R}^{k+n} be defined as follows:

\displaystyle{F(\mathbf{x},\mathbf{y})=\big(f(\mathbf{a}+\mathbf{y})-\mathbf{x},\pi_{n+1}(\mathbf{y}),\dots,\pi_{n+k}(\mathbf{y})\big),\quad\mathbf{x}\in\mathbf{R}^{n},\mathbf{y}\in\mathbf{R}^{k+n}}

then F(\mathbf{0}_n,\mathbf{0}_{n+k})=\mathbf{0}_{n+k}, and

\displaystyle{\frac{\partial F}{\partial \mathbf{y}}(\mathbf{0}_n,\mathbf{0}_{n+k})=\begin{bmatrix}D_1f(\mathbf{a})&\cdots&D_{n+1}f(\mathbf{a})&\cdots&D_{n+k}f(\mathbf{a})\\&&1&&\\&&&\ddots&\\&&&&1\end{bmatrix}}

this is an (n+k)\times (n+k) matrix which is non-singular, as we assume D_1f(\mathbf{a}),\dots,D_nf(\mathbf{a}) are linearly independent. Thus by the implicit function theorem ,there is a neighborhood B of \mathbf{0}_n in \mathbf{R}^n and a unique continuous function g:B\to\mathbf{R}^{k+n} such that g(\mathbf{0}_n)=\mathbf{0}_{n+k} and

\displaystyle{F(\mathbf{x},g(\mathbf{x}))=\mathbf{0},\quad\forall\mathbf{x}\in B}

Now if \mathbf{c} is close enough to \mathbf{0} such that \mathbf{c}\in B, we can have F(\mathbf{c},g(\mathbf{c}))=\mathbf{0}, which means f(\mathbf{a}+g(\mathbf{c}))=\mathbf{c}.
Finally, in the general case, for any n columns of Df(\mathbf{a}) which are linearly independent, which means there are k columns remaining, we substitute \pi_{n+1}(\mathbf{y}),\dots,\pi_{n+k}(\mathbf{y}) with the corresponding k column coordinate functions, and the conclusion follows.