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Contraction Mapping Method

Reference

Ordinary Differential Equation, Vladimir Igorevich Arnold

《常微分方程》 Л.С.庞特里亚金

压缩映射法 | Proof using Contraction Mapping

This part we want to prove Picard Theorem from another perspective. And this method is also universal on multi-dimensional Cauchy Problem. So to simplify the notation, we change the problem equivalently to

\[ \begin{equation} \dot{\mathbfit{x}} = \mathbfit{f}(t, \mathbfit{x}),\quad \mathbfit{x}(t_0) = \mathbfit{x}_0 \label{eq-vector-cauchy} \end{equation} \]

where \(\mathbfit{x}\in \mathbb{R}^n\) and \(\mathbfit{f}: \mathbb{R}^{n+1} \mapsto \mathbb{R}^n \in C(D)\) is a vector function(or vector field). To simplify the problem, we assume \(\pmb{f} \in C^r(D)\), where \(r\geq 1\).

Assume we give a Euvlid Structure in region \(D\in \mathbb{R}^{n+1}\). For all \((t_0,\pmb{x}_0)\in D\), we consider a cylinder(柱体) with sufficiently small parameters \(a\) and \(b\)

\[ \Gamma= \{(t, \pmb{x}): |t-t_0|\leq a, \|\pmb{x}-\pmb{x}_0\|\leq b\} \]

which is still a subset of \(D\).

Then the Picard Theorem becomes:

Picard Theorem

Assume \(\pmb{f}\) of problem \(\ref{eq-vector-cauchy}\) is continous and differentiable(or Lipschitz condition) on region \(\Gamma\), then for all given \(\pmb{x}\) which is sufficiently close to \(\pmb{x}_0\), there exsits a neighberhood of \(t_0\), such that there exists a unique solution \(\pmb{\varphi}(t)\).

  • Using the fixed point theorem (existence and uniqueness).

群论基础 | Group Basis

Firstly, let us introduce some basic ideas about groups.

群的定义 | Definition of Group

Firstly, we have to define Law of Composition.

Assume \(S\) is a set. A Law of Composition is a map

\[ S\times S \mapsto S. \]

\(S\times S\) deontes the product set, whose elements are pairs \(a\), \(b\) of elements of \(S\).

A group is a set \(G\) together with a law of composition(Here we use sign \(\cdot\) (multiplicative notation), and sign \(+\) (additive notation) also can be applied) that has the following properties:

  • The law of composition is associative.
\[ (ab)c=a(bc), \quad \forall a,b,c\in G \]
  • \(G\) contains an identity element \(1\) (\(0\) in sign \(+\)) such that
\[ 1a=a \text{ and } a1=a,\quad \forall a\in G \]
  • Every element \(a\) of \(G\) has an inverse, i.e. an element \(b\) such that
\[ ab=1 \text{ and } ba=1 \]

which is denoted by \(a^{-1}\) (\(-a\) in additive notation).

群的性质 | Properties of Group

  • Cancellation Law

Let \(a\), \(b\), \(c\) be elements of a group \(G\) whose law of composition is written multiplicatively. If

\[ ab=ac \text{ or } ba = ca \Rightarrow b=c. \]

If

\[ ab=a \text{ or } ba =a \Rightarrow b=1. \]

Multiply both sides of the above equation on the left by \(a^{-1}\).

Example.

\(n\times n\) general liear group is the group of all invertible \(n\times n\) matrices, denoted by

\[ GL_{n} =\{n\times n \text{ invertible matrices } A\}. \]

where the law of composition is matrix multiplication.

同态 | Homomorphism

Let \(G\) and \(G'\) be groups written with multiplicative notation. A Homomorphism \(\varphi:G\mapsto G'\) is a map from \(G\) to \(G'\) such that

\[ \varphi(ab) = \varphi(a)\varphi(b), \quad \forall a,b\in G \]

If we let \(\varphi\) to be a bijection, then we call it Isomorphism(同构).

Example.

the determinent function det: \(GL_n(\mathbb{R})\mapsto \mathbb{R}^\times\)

同态的性质 | Properties of Homomorphism

Let \(\varphi: G\mapsto G'\) be a group homomorphism.

(i) If \(a_1,a_2,\cdots, a_k\) are elements of \(G\), then

\[ \varphi(a_1 a_2\cdots a_k)=\varphi(a_1)\varphi(a_2)\cdots\varphi(a_n) \]

(ii) \(\varphi\) maps the identity to identity, i.e.

\[ \varphi(1_G)=1_{G'} \]

(iii) \(\varphi\) maps inverses to inverses, i.e.

\[ \varphi(a^{-1}) = \varphi(a)^{-1} \]

(i) by induction(Strong induction).

(ii) using \(1\cdot 1=1\) and \(\varphi(1)\varphi(1)=\varphi(1\cdot 1)=\varphi(1)\), cancel both sides to obtain and get \(\varphi(1)=1_{G'}\).

(iii) similarly, \(a^{-1}a=1\) and \(\varphi(a^{-1})\varphi(a)=\varphi(a^{-1}a)=\varphi(1_G)=1_{G'}\), and we are done.

相空间 | Phase Space

Then we have to introduce some comception about phase.

单参数变换群 | One-Parameter Group of Transformation

\(M\) is a set. A family of mapping \(\{g^t\}_{t\in \mathbb{R}}\) which maps set \(M\) into itself is called One-Parameter Group of Transformation of set \(M\), if

\[ g^{t+s} = g^tg^s, \quad \forall t,s\in \mathbb{R} \]

and \(g^0\) is identity mapping(恒等映射).

相流、相空间、运动、相曲线的定义 | Definition of Phase Flow

A couple composed of set \(M\) and its one-parameter group of transformation \(\{g^t\}\), denoted as \((M, \{g^t\})\), is called Phase Flow. And here \(M\) is called the phase space of the phase flow. The element of \(M\) is called Phase Point.

Consider a mapping

\[ \begin{equation} \varphi: \mathbb{R}\mapsto M, \quad \varphi(t) = g^tx,\quad x\in M \label{map-motion} \end{equation} \]

which maps a real straight line into phase space. Then it is called a motion of phase point \(x\) under the action of phase flow.

The image of \(\mathbb{R}\) under mapping \(\varphi\) is called the phase curve of phase flow \((M,\{g^t\})\).

  • Fixed Point

If the phase curve of a phase point \(x\in M\) is itself, i.e.

\[ g^tx=x,\quad \forall t\in \mathbb{R} \]

then \(x\) is called the fixed point of the phase flow \((M, \{g^t\})\).

In fact one-parameter group of transformation is exchangeable(\(g^tg^s=g^{t+s}=g^{s+t}=g^sg^t\)). And it is also a bijection. This is easy to prove.

Firstly we prove it is surjection. \(\forall x\in M\), \(\exists g^{-t}x \in M\), such that \(g^t(g^{-t}x)=x\). Then we prove it is a injective mapping. \(g^tx=g^ty\), then \(x=g^0x=g^{-t}g^tx=g^{-t}g^ty=g^0y=y\).

With the above property we can easily see the following theorem.

相空间中的点仅有一条相曲线

For all \(x\in M\), there only exists one phase curve.

Because \(g^t\) is a bijection.

Now we introduce two important conceptions.

映射的图形 | Graph of a mapping

The graph of a mapping \(f : A\mapsto B\) is a subset of the direct product(直积) \(A\times B\):

\[ \{(a,f(a))| a\in A\} \]

扩张相空间、积分曲线 | Expanded Phase Space, Integral Curve

The Expanded Phase Space of phase flow \((M, \{g^t\})\) is the direct product \(\mathbb{R} \times M\).

The Integral Curve of phase flow \((M, \{g^t\})\) is the graph of the motion \(\ref{map-motion}\).

Now we have to make use of DIfferential in Euclid Space.

可微函数、可微映射、微分同胚 | Differentiable Function, Differentiable Mapping, Diffeomorphism

Assume \(U \subset\mathbb{R}^n, V \subset\mathbb{R}^m\). Then

A Differentiable Function is a function \(f: U\mapsto \mathbb{R}\) which is \(r\) times differentiable.

A Differentiable Mapping is a mapping \(f:U\mapsto V\) defined by

\[ y_i = f_i(x_1,x_2,\cdots,x_n), \quad i=1,2\cdots, m \]

where \(f_i: U\mapsto \mathbb{R}\) is a Differentiable function. If \(y_i: V\mapsto \mathbb{R}\) is a coordinate of \(\pmb{y}\in \mathbb{R}^m\), then \(y_i\circ f: U\mapsto \mathbb{R}\) is a Differentiable function in \(U\).

A Diffeomorphism is a bijection \(f:U\mapsto V\), such that \(f\) and \(f^{-1}\) are both Differentiable mappings.

与坐标轴有关的相速度、向量场 | Phase Speed, Vector Field

The Phase Speed \(\pmb{v}(x)\) of phase flow \(g^t\) at point \(x\in M\) is

\[ \pmb{v}(x) = \frac{d}{dt}\Bigg|_{t=0}g^tx \]

And at time \(\tau\), we have phase speed

\[ \pmb{v}(g^\tau x) = \frac{d}{dt}\Bigg|_{t=\tau}g^tx. \]

Now we let \(M\) to be a region in Euclid Space \(\mathbb{R}^n\) with coordinates \(x_1,x_2,\cdots,x_n\). And if \(x_i:M\mapsto \mathbb{R}\) is the coordinate of \(\pmb{x}\in M\), then the vector \(\pmb{v}(x)\) is defined by \(v_i: M\mapsto \mathbb{R}, i=1,2\cdots, n\):

\[ v_i(\pmb{x}) = \frac{d}{dt}\Bigg|_{t=0}x_i(g^t\pmb{x}) \]

So we define a Vector Field \(\pmb{v}\) on \(M\) when neglecting \(t\).

与坐标轴无关的相速度、向量场 | Phase Speed, Vector Field

Firstly, we have to define that a coordinate \(\{y_i\}_{i=1}^{n}:U\mapsto \mathbb{R}\) is admissive, if mapping

\[ y:U\mapsto \mathbb{R}^n,\quad y(\pmb{x}) = y_1(\pmb{x})\pmb{e}_1 + y_2(\pmb{x})\pmb{e}_2 + \cdots +y_n(\pmb{x})\pmb{e}_n \]

is a diffeomorphism.

So we have a proposition:

Two curves \(\varphi_1\), \(\varphi_2\) that pass \(\pmb{x}\in U\) are tangent with each other if and only if two curves \(y\circ \varphi_1\), \(y\circ\varphi_2\) that pass \(y(\pmb{x})\in V\) are tangent with each other.

So the velocity vector of curve \(\varphi:I\mapsto U\) that pass \(\pmb{x}\in U\) is

\[ \pmb{v}=\dot{\varphi}(0), \quad \pmb{v}=\frac{d\varphi}{dt}\Bigg|_{t=0}. \]

切向量、切空间 | Tangent Vector, Tangent Space

Assume \(U\in \mathbb{R}^n\) with coordinates \(x_1,x_2,\cdots, n\) and map \(\varphi:I\mapsto U\) maps an interval on \(\mathbb{R}\) to \(U\) such that \(\varphi(0)=\pmb{x}\in U\). and also its velocity is determined by

\[ v_i = \frac{d}{dt}\Bigg|_{t=0}(x_i\circ \varphi), i=1,2,\cdots, n. \]

Two curves \(\varphi_1, \varphi_2:I\mapsto U\) pass the same point \(\pmb{x}\) are tangent with each other if \(t\rightarrow 0\), \(\rho(\varphi_1, \varphi_2)\rightarrow 0\).

A set composed by all tangent vectors of the curves passing \(\pmb{x}\) is a linear space with dimension \(n\), which is called Tangent Space, denoted by \(TU_x\).

Now we want to give a space without coordinates.

映射的导数 | Derivative of Mapping

Assume that \(f:U\mapsto V\) is a differentiable mapping from \(\pmb{x}=(x_1,x_2,\cdots,x_n)\) to \(\pmb{y}=(y_1,y_2,\cdots,y_m)\). Let \(\pmb{y}=f(\pmb{x})\in V\).

The Derivative of mapping \(f\) at \(\pmb{x}\) is a mapping from tangent space at \(\pmb{x}\in U\) to tangent space at \(\pmb{y}\in V\)

\[ f_*|_{\pmb{x}}: TU_{\pmb{x}} \mapsto TV|_{f(\pmb{x})} \]

This mapping maps velocity vector of curve \(\varphi\) at \(\pmb{x}\) into velocity vector of \(f\circ \varphi\) at \(f(\pmb{x})\), i.e.

\[ f_*|_{\pmb{x}}\left(\frac{d\varphi}{dt}\Bigg|_{t=0} \right) = \frac{d}{dt}\Bigg|_{t=0}(f\circ \varphi). \]

which defines a linear mapping from \(TU_{\pmb{x}}\) to \(TV_{f(\pmb{x})}\).

度量空间下的压缩映射 | Contraction Mapping in Metric Space

Then, let us recall the definition of metric space(度量空间).

度量空间的定义 | Definition of Metric Space

Assume \(M\) is a set. If there is a function \(\rho: M\times M\mapsto \mathbb{R}\) defined on it, satisfies that, \(\forall x,y,z \in M\)

  • Positive
\[ \rho(x,y)>0,\quad x\neq y \]
  • Definite
\[ \rho(x, y)=0 \Leftrightarrow x=y \]
  • Symmetry
\[ \rho(x,y) = \rho(y,x) \]
  • Triangle inequality
\[ \rho(x,z) \leq \rho(x,y) +\rho(y,z) \]

then \(M\) is called metric space with metrc \(\rho\).

Now we can rewrite the Lipschitz Condition with terms of metric space.

Lipschitz 条件 | Lipschitz Condition

Assume \(A: M_1 \mapsto M_2\) is a mapping from a metric space \(M_1\) (with metric \(\rho_1\)) to another matric space \(M_2\) (with metric \(\rho_1\)). If there exists a constant \(L>0\), s.t.

\[ \rho_2(Ax,Ay)\leq L\rho_1(x,y),\quad \forall x,y \in M_1 \]

then we say mapping \(A\) satisfies Lipschitz Condition.

In Proof 1 we will show how to use Lipschitz condition to shrink the region of \((t, \pmb{x})\)(Expanded Phase Space) such that the contraction mapping maps the metric space into itself.

Usually, a mapping that maps a space \(M\) into itself would be of great use for us, so here comes the following definition.

压缩映射的定义 | Definition of Contraction mapping

Assume \(A: M \mapsto M\) is a mapping from complete metric space with metric \(\rho\) to itself. If there exists a constant \(k\), \(0<k<1\), s.t.

\[ \rho(Ax,Ay)\leq k\rho(x, y),\quad \forall x, y\in M. \]

then \(A\) is called a Contraction mapping.

If \(Ax =x \in M\), we call \(x\) is a fixed point of mapping \(A\). Then the following theorem is quite useful in our future proof.

Banach不动点定理 | Banach Fixed-Point Theorem

(This also known as contraction mapping theorem.)

Assume \(A: M\mapsto M\) is a contraction mapping, then \(A\) has a unique fixed-point.

We prove by showing that \(A^nx\) is a Cauchy Sequence and its limit \(X\) falls in \(M\). Then use the property of limits to show it is fixed-point of \(A\). Then prove the uniqueness by using property of metric \(\rho\).

Let \(d = \rho(x,Ax)\), then

\[ \rho(A^nx,A^{n+1}x)\leq k^n\rho(x,Ax) = k^nd \]

Sequence \(A^nx\), \(n=1,2,\cdots\) is convergent. So its limit exists and let \(X \overset{\Delta}{=} \lim\limits_{n\rightarrow \infty}A^nx\).

So

\[ AX = A \lim\limits_{n\rightarrow \infty}A^nx = \lim\limits_{n\rightarrow \infty}A^{n+1}x = X \]

which means \(X\) is a fixed-point of \(A\). Then we prove the uniqueness of fixed point. Assume there are two fixed points of \(A\), denoted as \(X\), \(Y\). See that

\[ \rho(X, Y) =\rho(AX, AY) \leq k\rho(X,Y) \]

Because \(0<k<1\), so \(\rho(X,Y)=0\) \(\Rightarrow\) \(X=Y\).

Readers can compare this theorem to Fixed-Point Theorem in Numerical Analysis.

Note that we have an abstract set \(M\) with abstract metric \(\rho\). In the next two parts, we will let \(M\) to be a set of mappings and \(\rho\) to be defined based on natural norm of vectors.

证明1 | Proof 1

Now we consider Picard mapping \(A\) is a mapping from a mapping to another mapping

\[ A: (\pmb{\varphi}: t\mapsto \pmb{x}) \mapsto (A\pmb{\varphi}: t\mapsto \pmb{x}) \]

defined by

\[ (A\pmb{\varphi})(t) = \pmb{x}_0 + \int_{t_0}^{t}\pmb{f}(\tau, \pmb{\varphi}(\tau))d\tau \]

Geometrically speaking, tangent line of each point on \(A\pmb{\varphi}\), i.e. \((A\pmb{\varphi})'(t)\), equals to the vector field determined by \(\pmb{\varphi}(t)\), i.e. \(f(t, \pmb{\varphi}(t))\). This can help us find a smaller region for \(A\pmb{\varphi}\) to be well-defined.

And note that \(\pmb{\varphi}\) is a solution of Cauchy Problem \(\ref{eq-vector-cauchy}\) if and only if \(\pmb{\varphi} = A \pmb{\varphi}\). So we have to prove \(A\) is a contraction mapping in a complete metric space.

  • 定义域\((t, \pmb{x})\) | Some Constants \(a\), \(b\), \(M\)

Let

\[ M=\max_{\pmb{x}\in \Gamma}\|\pmb{f}\|, \quad L= \max_{\pmb{x}\in \Gamma}\|\pmb{f}_*\| \text{(or by Lipschitz condition)} \]

which can be obtained because \(\Gamma\) is a compact set. Firstly, we consider a Cylinder

\[ \Gamma = \{(t,\pmb{x}): |t-t_0|\leq a, \|\pmb{x}-\pmb{x}_0\|\leq b\} \]

Now consider a Cone(锥体) with opening \(M\) and sufficient small height \(a\) at point \((t_0,\pmb{x}_0)\)

\[ K_0 = \{(t,\pmb{x}): |t-t_0|\leq a, \|\pmb{x}-\pmb{x}_0\|\leq M|t-t_0|\} \]

Which means our mapping \(\pmb{\varphi}\) after contraction mapping \(A\) is still well-defined. See details in the following proof.

\[ \begin{align*} \|A\pmb{\varphi}(t) - \pmb{x}_0\| &=\left\|\int_{t_0}^t\pmb{f}(\tau,\pmb{\varphi}(\tau))d\tau\right\| \\ &\leq \left|\int_{t_0}^t\|\pmb{f}(\tau,\pmb{\varphi}(\tau))\|d\tau\right|\\ &\leq M |t-t_0| \end{align*} \]

So as long as \(a<b/M\), \(\|A\pmb{\varphi}(t)-\pmb{x}_0\|\leq b\).

  • Prove \(\pmb{\varphi}\) compose a complete metric space.

We define a Space \(M\) composed of mappings \(\pmb{\varphi}: \mathbb{R}\mapsto \mathbb{R}^n\) defined by \(\pmb{\varphi}(t)=\pmb{x}\).

Define a metric \(\rho\) to be

\[ \rho(\pmb{\varphi}_1, \pmb{\varphi}_2) = \|\pmb{\varphi}_1-\pmb{\varphi}_2\| = \max_{|t-t_0|\leq a}|\pmb{\varphi}_1(t)-\pmb{\varphi}_2(t)| \]

Then \(\{M, \rho\}\) is a complete metric space.

  • Prove \(A\) is a contraction mapping.

That is, \(\forall \pmb{\varphi}_1, \pmb{\varphi}_2 \in M\)

\[ \begin{align*} \|A\pmb{\varphi}_1- A\pmb{\varphi}_2\| &= \left\|\int_{t_0}^t[\pmb{f}(\tau, \pmb{\varphi}_1(\tau))-\pmb{f}(\tau, \pmb{\varphi}_2(\tau))]d\tau\right\|\\ &\leq \left|\int_{t_0}^t\|\pmb{f}(\tau, \pmb{\varphi}_1(\tau))-\pmb{f}(\tau, \pmb{\varphi}_2(\tau))\|d\tau \right|\\ &\leq L \int_{t_0}^t \|\pmb{\varphi}_1(\tau) - \pmb{\varphi}_2(\tau)\|d\tau \\ &\leq L \|\pmb{\varphi}_1 - \pmb{\varphi}_2\| \int_{t_0}^t d\tau = L |t-t_0| \|\pmb{\varphi}_1 - \pmb{\varphi}_2\| \end{align*} \]

So as long as \(a<1/L\), \(A\) is a contraction mapping.

By Banach Fixed Point Theorem, there exists only one soluion of ODE.

证明2 | Proof 2

In this part, we consider a different mapping.

  • 定义域\((t, \pmb{x})\)

In order to let all the possible cone to be in the cylinder, we have to shrink the region to make the region is well defined. That is, there exsits sufficiently small parameters \(b'<b\) such that \(\|\pmb{x}-\pmb{x}_0\|<b'\). So now we get a smaller region (a smaller cylinder)

\[ \Gamma' = \{(t, \pmb{x}): |t-t_0|<a', \|\pmb{x}-\pmb{x}_0\|<b'\} \]

We also have to prove that \(A\) maps \(M\) into itself. That is, \(\|\varphi(t)-\pmb{x}_0\|\leq b'\). Here we have to make \(a'\) to satisfy some condition.

Consider all the possible continuous mapping \(\pmb{h}\) which maps the above cylinder into \(\mathbb{R}^n\), which is a solution of original ODE problem. Assume \(M\) is a set composed of these mappings with additional condition

\[ \|\pmb{h}(t, \pmb{x})\| \leq C|t-t_0| \]

Specially, we let \(\pmb{h}(t_0,\pmb{x})=0\). We introduce a metric

\[ \rho(\pmb{h}_1,\pmb{h}_2)=\|\pmb{h}_1-\pmb{h}_2\|=\max_{(t,\pmb{x})\in \Gamma' }\|\pmb{h}_1(t,\pmb{x})-\pmb{h}_2(t,\pmb{x})\| \]

Note: Space \(M\) is dependent on \(a'\), \(b'\) and \(C\).

Now we really have to consider a mapping

\[ A: M\mapsto M \]

defined by

\[ (A\pmb{h})(t, \pmb{x}) = \int_{t_0}^t\pmb{f}(\tau, \pmb{x}+\pmb{h}(\tau, \pmb{x}))d\tau \]

\(A\) 是压缩映射 | \(A\) is a contraction mapping

\(A\) is a contraction mapping from \(M\) to \(M\), if \(a'\) is sufficiently small.

Prove it. The following proof are set to find how small \(a'\) shoulc be.

Firstly, prove \(A\) maps \(M\) to itself. Then Prove if is a constraction mapping.

  • \(A\) maps \(M\) to itself.
\[ \begin{align*} \|(A\pmb{h}(t, \pmb{x}))\| &\leq \|\int_{t_0}^t\pmb{f}(\tau, \pmb{x}+\pmb{h}(\tau,\pmb{x}))d\tau\| \\ &\leq |\int_{t_0}^tCd\tau| \leq C|t-t_0| \end{align*} \]

so \(AM\subset M\).

  • \(A\) is a constraction mapping.

Estimate \(A\pmb{h}_1- A\pmb{h}_2\):

\[ (A\pmb{h}_1-A\pmb{h}_2)(t,\pmb{x}) = \int_{t_0}^t[\pmb{f}(\tau, \pmb{x}+\pmb{h}_1(\tau,\pmb{x}))-\pmb{f}(\tau, \pmb{x}+\pmb{h}_2(\tau,\pmb{x}))]d\tau \]

Denote \(\pmb{f}_i(\tau) = \pmb{f}(\tau, \pmb{x}+\pmb{h}_i(\tau,\pmb{x}))\), \(i=1,2\), so

\[ \begin{align*} \|\pmb{f}_1(\tau)- \pmb{f}_2(\tau)\| &\leq L\|\pmb{h}_1(\tau)-\pmb{h}_2(\tau)\| \quad\text{(using Lipschitz condition)}\\ &\leq L\|\pmb{h}_1-\pmb{h}_2\| = L\rho(\pmb{h}_1, \pmb{h}_2) \quad \text{(by definition of metric)} \end{align*} \]

And

\[ \|(A\pmb{h}_1-A\pmb{h}_2)(t,\pmb{x})\| \leq L a'\rho(\pmb{h}_1, \pmb{h}_2) \]

So if \(La'<1\), then \(A\) is a contraction mapping.

附录：连续可微到Lipschitz连续 | Appendix: from Continuity & Differentiability to Lipschitz Condition

we will choose natural metric

\[ \rho(\pmb{x}, \pmb{y}) = \|\pmb{x}-\pmb{y}\| = \sqrt{(\pmb{x}-\pmb{y},\pmb{x}-\pmb{y})} \]

so space \(\mathbb{R}^n\) with the above metric is a complete metric space.

So in this case, we have a parallel theorem for smoothness and Lipschitz condition in \(\mathbb{R}\).

连续可微映射满足Lipschitz条件 | Continuously Differentiable mapping satisfies Lipschitz condition

Assume \(V\subset U \subset\mathbb{R}^m\) is a convex and contract set, and continuously differentiable mapping \(\pmb{f}\) which maps \(U\) to \(\mathbb{R}^n\) satisfies Lipschitz condition, and the constant

\[ L = \sup_{\pmb{x}\in V} \|\pmb{f}_{*\pmb{x}}\| \]

where \(\pmb{f}_*|_{\pmb{x}}=\pmb{f}_{*\pmb{x}}:T\mathbb{R}^m_{\pmb{x}}\mapsto T\mathbb{R}^n_{\pmb{x}}\)

Prove it.

Let \(\pmb{z}(t)=\pmb{x}+t(\pmb{y}-\pmb{x})\), \(0\leq t\leq 1\). Then

\[ \begin{align*} \pmb{f}(\pmb{y})-\pmb{f}(\pmb{x})&=\int_{0}^1\frac{d}{dt}\pmb{f}(\pmb{z}(\tau))d\tau\\ &=\int_0^1\pmb{f}_{*\pmb{z}(\tau)}[\pmb{\dot{z}}(\tau)]d\tau \\ &=\int_0^1 \pmb{f}_{*\pmb{z}(\tau)} (\pmb{y}-\pmb{x})d\tau \end{align*}\\ \]

where \(\pmb{y}-\pmb{x}\) is a constant, so

\[ \begin{align*} \left\| \pmb{f}(\pmb{y})-\pmb{f}(\pmb{x}) \right\| &=\left\|\int_0^1 \pmb{f}_{*\pmb{z}(\tau)} (\pmb{y}-\pmb{x})(\tau)d\tau\right\|\\ &\leq \int_0^1 \|\pmb{f}_{*\pmb{x}}\| \|\pmb{y}-\pmb{x}\|d\tau\\ &\leq L\|\pmb{y}-\pmb{x}\| \end{align*} \]

and we are done.