Analysis on Manifolds 8. The Inverse Function Theorem

christangdt

这一节和下一节讨论的是两个比较特殊的多元微分情景，即反函数和隐函数，本节的主要结论是只要导数Df(x)是非奇异的，那么f有一个可微的反函数g。作者将过程分成了三个定理逐步证明。

Exercises

Exercise 1. Let f:\mathbf{R}^2\to\mathbf{R}^2 be defined by the equation f(x,y)=(x^2-y^2,2xy).
( a ) Show that f is one-to-one on the set A consisting of all (x,y) with x>0.
( b ) What is the set B=f(A)?
( c ) If g is the inverse function, find Dg(0,1).
Solution:
( a ) Suppose f(x,y)=f(a,b), in which x>0,a>0, then we know x^2-y^2=a^2-b^2 and xy=ab, also since |f(x,y)|=|f(a,b)|, we have (x^2-y^2 )^2+4x^2 y^2=(a^2-b^2 )^2+4a^2 b^2, thus x^2+y^2=a^2+b^2, together with x^2-y^2=a^2-b^2 we have x^2=a^2, and thus x=a, together with xy=ab we get y=b, so f is one-to-one.
( b ) Since A={(x,y):x>0}, we write x=a \cos\theta,y=a \sin\theta, in which a>0,\theta\in(-\pi/2,\pi/2), then f(a,\theta)=(a^2 \cos2\theta,a^2\sin2\theta ), in which 2\theta\in(-\pi,\pi), thus we can never have \cos2\theta=-1, also \cos2\theta and \sin2\theta cannot be 0 simultaneously, we get B=\mathbf{R}^2-{(x,0):x\leq0}.
( c ) We have Dg(0,1)=[Df(g(0,1))]^{-1}, in which g(0,1)=(x,y), which satisfies f(x,y)=(0,1), solve it we get x=y=1/\sqrt{2}, as

\displaystyle{Df(x,y)=\begin{bmatrix}2x&2y\\-2y&2x\end{bmatrix},\quad Df\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=\begin{bmatrix}\sqrt{2}&\sqrt{2}\\-\sqrt{2}&\sqrt{2}\end{bmatrix},\quad \left[Df\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right]^{-1}=\frac{1}{4}\begin{bmatrix}\sqrt{2}&-\sqrt{2}\\{\sqrt{2}}&\sqrt{2}\end{bmatrix}}

we know that

\displaystyle{Dg(0,1)=\frac{1}{4}\begin{bmatrix}\sqrt{2}&-\sqrt{2}\\{\sqrt{2}}&\sqrt{2}\end{bmatrix}}

Exercise 2. Let f:\mathbf{R}^2\to\mathbf{R}^2 be defined by the equation f(x,y)=(e^x\cos{y},e^x\sin{y}).
( a ) Show that f is one-to-one on the set A consisting of all (x,y) with 0<y<2\pi.
( b ) What is the set B=f(A)?
( c ) If g is the inverse function, find Dg(0,1).
Solution:
( a ) Suppose f(x,y)=f(a,b), in which 0<y<2\pi,0<b<2\pi, then since |f(x,y)|=|f(a,b)|, we have e^{2x}=e^{2a}, thus x=a, so we have \cos{y}=\cos{b} and \sin{y}=\sin{b}, thus y=b.
( b ) Since A={(x,y):0<y<2\pi}, we know \cos{y}\neq 1, and when \sin{y}=0,\cos{y}=-1, thus B=\mathbf{R}^2-{(0,y):y\leq 0}.
( c ) We have Dg(0,1)=[Df(g(0,1))]^{-1}, in which g(0,1)=(x,y), which satisfies f(x,y)=(0,1), solve it we get x=0,y=\pi/2, as

\displaystyle{Df(x,y)=\begin{bmatrix}e^x\cos{y}&e^x\sin{y}\\-e^x\sin{y}&e^x\cos{y}\end{bmatrix},\quad Df\left(0,\frac{\pi}{2}\right)=\begin{bmatrix}0&1\\-1&0\end{bmatrix}}

we have

\displaystyle{Dg(0,1)=\left[Df\left(0,\frac{\pi}{2}\right)\right]^{-1}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}}

Exercise 3. Let f:\mathbf{R}^n\to\mathbf{R}^n be given by the equation f(\mathbf{x})=|\mathbf{x}|^2\cdot\mathbf{x}. Show that f is of class C^{\infty} and that f carries the unit ball B(\mathbf{0};1) onto itself in a one-to-one fashion. Show, however, that the inverse function is not differentiable at \mathbf{0}.
Solution: That f is of class C^{\infty} is easy to verify. Next we show f carries B(\mathbf{0};1) onto itself in a one-to-one fashion.
Suppose f(\mathbf{x})=f(\mathbf{y}) with \mathbf{x},\mathbf{y}\in B(\mathbf{0};1), write \mathbf{x}=(x_1,\dots,x_n),\mathbf{y}=(y_1,\dots,y_n), then we have

\displaystyle{(x_1^2+\cdots +x_n^2)x_j=(y_1^2+\cdots +y_n^2)y_j,\quad j=1,\dots,n}

When all x_j and y_j are 0, the result holds naturally, if any x_k\neq 0, then the corresponding y_k\neq 0 and we have

\displaystyle{\frac{x_k}{y_k}=\frac{y_1^2+\cdots +y_n^2}{x_1^2+\cdots +x_n^2}}

Let x_k/y_k=C, then we have x_k=Cy_k and x_j=Cy_j for j=1,\dots,n, thus

\displaystyle{(C^2y_1^2+\cdots +C^2y_n^2)\cdot (Cy_1,\dots,Cy_n)=(y_1^2+\cdots +y_n^2)\cdot (y_1,\dots,y_n)}

which means C^3=1 and C=1, so \mathbf{x}=\mathbf{y}.
Conversely, for any \mathbf{y}\in B(\mathbf{0};1), we can write \mathbf{y}=(y_1,\dots,y_n), then if \mathbf{y}\neq \mathbf{0} we let \mathbf{x}=(x_1,\dots,x_n) with

\displaystyle{x_j=\frac{y_j}{y_1^2+\cdots+y_n^2},\quad j=1,\dots,n}

we have f(\mathbf{x})=\mathbf{y}. If \mathbf{y}=\mathbf{0} we just notice that f(\mathbf{0})=\mathbf{0}.
Finnaly, we show the inverse function is not differentiable at \mathbf{0}. Assume so, then Df(\mathbf{0}) shall be non-singular. But it is easy to calculate D_jf_i(\mathbf{0})=0 for i,j=1,\dots,n, thus Df(\mathbf{0}) is the square matrix with all entries 0, and singular, this is a contradiction.

Exercise 4. Let g:\mathbf{R}^2\to\mathbf{R}^2 be given by the equation g(x,y)=(2ye^{2x},xe^y). Let f:\mathbf{R}^2\to\mathbf{R}^3 be given by the equation f(x,y)=(2x-y^2,2x+y,xy+y^3).
( a ) Show that there is a neighborhood of (0,1) that g carries in a one-to-one fashion onto a neighborhood of (2,0).
( b ) Find D(f\circ g^{-1}) at (2,0).
Solution:
( a ) By Lemma 8.1, it is enough to to show that Dg(0,1) is non-singular, since

\displaystyle{Dg(x,y)=\begin{bmatrix}4ye^{2x}&2e^{2x}\\e^y&xe^y\end{bmatrix}}

we have

\displaystyle{Dg(0,1)=\begin{bmatrix}4&2\\e&0\end{bmatrix},\quad \det{Dg(0,1)}=-2e}

( b ) First, g^{-1}(2,0)=(0,1) and f(0,1)=(-1,1,1). By the chain rule we have

\displaystyle{D(f\circ g^{-1})(2,0)=Df(0,1)\cdot Dg^{-1}(2,0)=Df(0,1)\cdot [Dg(0,1)]^{-1}}

We calculate Df(0,1):

\displaystyle{Df(x,y)=\begin{bmatrix}2&-2y\\2&1\\y&x+3y^2\end{bmatrix},Df(0,1)=\begin{bmatrix}2&-2\\2&1\\1&3\end{bmatrix}}

also we have

\displaystyle{[Dg(0,1)]^{-1}=\frac{1}{2e}\begin{bmatrix}0&2\\e&-4\end{bmatrix}}

so

\displaystyle{D(f\circ g^{-1})(2,0)=\frac{1}{2e}\begin{bmatrix}2&-2\\2&1\\1&3\end{bmatrix}\begin{bmatrix}0&2\\e&-4\end{bmatrix}=\frac{1}{2e}\begin{bmatrix}-2e&-4\\e&0\\3e&-10\end{bmatrix}}

Exercise 5. Let A be open in \mathbf{R}^n; let f:A\to\mathbf{R}^n be of class C^r; assume Df(x) is non-singular for \mathbf{x}\in A. Show that even if f is not one-to-one on A, the set B=f(A) is open in \mathbf{R}^n.
Solution: Choose any \mathbf{b}\in\ B, since B=f(A), there exists \mathbf{a}\in A such that f(\mathbf{a})=\mathbf{b}, since Df(\mathbf{a}) is non-singular, by Theorem 8.3, there is a neighborhood U of \mathbf{a} such that V=f(U) is open in \mathbf{R}^n and f is an isomorphism on U, the inverse function g=f^{-1} is of class C^r.
Since A is open, we can find a neighborhood U' of \mathbf{a} such that U'\subset A, the set U\cap U' is a neighborhood of \mathbf{a} and is contained in A, which means f(U\cap U') is contained in B. Since g is continuous and f(U\cap U')=g^{-1}(U\cap U'), we see that f(U\cap U') is open, the conclusion follows as \mathbf{b}\in\ f(U\cap U').

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