Metric Spaces §

Introduction and Common Metrics §

A set $X$ and a function of two variables $d:X\times X\to \mathbb{R}$ is a metric space if for for all $u,v,w\in X$, the following 3 properties are true:

1. Non-Negativity: $d(u,v)\ge 0$, and $d(u,v)=0$ if and only if $u=v$
2. Symmetry: $d(u,v)=d(v,u)$, $\forall u,v\in X$
3. Triangle Inequality: $d(u,v)\le d(u,w)+d(w,v)$, $\forall u,v,w\in X$.

In this case, we call $d$ a metric on $X$. Note that a single set could have multiple valid metrics, though there are some metrics that are more commonly used than others.

Note that if $Y$ is a metric space with metric $d$, then any $X\subseteq Y$ is also a metric space with metric $d$.

We describe 3 of the most important metric spaces in the below theorems.

Theorem: Common Metric on $\mathbb{R}$

For any two real numbers $p,q\in \mathbb{R}$, we have the metric on $\mathbb{R}$

$$d(p,q)=|p-q|$$

The proof for the metric is trivial, as it all follows from properties of absolute values.

Theorem: Common Metric on ${\mathbb{R}}^n$

For any two points $p,q\in {\mathbb{R}}^n$, we have the metric on ${\mathbb{R}}^n$, otherwise known as the norm,

$$d(p,q)=||p-q||=\sqrt{\sum _{i=1}^n(p_i-q_i{)}^2}$$

Like the prior metric, the properties for the norm can be proven fairly easily.

Example: Norms (1)

In ${\mathbb{R}}^2$, some of the following are examples of norms.

We have the typical norm we are used to, $||x||=\sqrt{x_1^2+x_2^2}$.

But we also have other options for norms! For example,

$$\begin{array}{r}||x|{|}_1=|x_1|+|x_2|\\ ||x|{|}_{\infty }=\max (|x_1|,|x_2|)\end{array}$$

The proof for the former is trivial. For (3) of the latter, note that

$$\begin{array}{rl}||x+y|{|}_{\infty }& =\max \{|x_1+y_1|,|x_2+y_2|\}\\ & \le \max \{|x_1|+|y_1|,|x_2|+|y_2|\}\\ & \le \max \{|x_1|,|x_2|\}+\max \{|y_1|,|y_2|\}\end{array}$$

And comparing the norms, we find that

$$\max \{|x_1|,|x_2|\}\le \sqrt{x_1^2+x_2^2}\le |x_1|+|x_2|\le 2\max \{|x_1|,|x_2|\}$$

So in finite dimensions, we find that convergence along any one of the norms implies convergence along the others, by squeeze!

This fails to hold in infinite dimenions!

Theorem: Common Metric on $C([a,b],\mathbb{R})$

Let $C([a,b],\mathbb{R})$ be the set of all continuous functions $f:[a,b]\to \mathbb{R}$.

For any two functions $f,g\in C([a,b],\mathbb{R}$, we have the metric

$$d(f,g)=\underset{x\in [a,b]}{\max }|f(x)-g(x)|$$

This finds the maximum single difference between the functions in the interval $[a,b]$!

Remark

In the metric space $C([0,1],\mathbb{R})$, with norm

$$||f||=\underset{x\in [0,1]}{\max }|f(x)|$$

We find that closed and bounded sets need not be sequentially compact.

For example, consider the set $B=\{f:||f||\le 1\}$, and take sequence $f_k(x)=x^k$ in $B$. Then,

$$f_k(x)=\{\begin{array}{ll}0& 0\le x<1\\ 1& x=1\end{array}$$

Converges pointwise. However, there does not exist a convergent $f_{k_l}$!

Example: Norms (2)

Let $X$ be a set of continuous functions on $[a,b]$ to $\mathbb{R}$.

$$x=C([a,b],\mathbb{R})=\{f:[a,b]\to \mathbb{R},f\text{ continuous}\}$$

We can define the following norms and verify them. Note that we only verify their triangle inequalities, as the rest is obvious to show.

We define the norm $||f||=||f|{|}_{\infty }={\max }_{x\in [a,b]}|f(x)|$.

$$\begin{array}{r}\underset{x\in [a,b]}{\max }|f_1(x)+f_2(x)|\le \underset{x\in [a,b]}{\max }|f_1(x)|+\underset{x\in [a,b]}{\max }|f_2(x)|\end{array}$$

We can also define the norm $||f|{|}_1={\int }_a^b|f(x)|dx$.

$${\int }_a^b|f_1(x)+f_2(x)|dx\le {\int }_a^b|f_1(x)|dx+{\int }_a^b|f_2(x)|dx$$

We can also define $||f|{|}_2={\left({\int }_a^b{f}^2(x)dx\right)}^{1/2}$.

$$\begin{array}{rl}{\left({\int }_a^b(f_1(x)+f_2(x){)}^2\right)}^{1/2}& ={\left({\int }_a^b{f}_1(x{)}^2+{\int }_a^{b}2f_1(x)f_2(x)+{\int }_a^b{f}_2(x{)}^2\right)}^{1/2}\\ & \le {\left({\int }_a^b{f}_1(x{)}^2+2{\int }_a^b(f_1^2(x){)}^{1/2}(f_2^2(x){)}^{1/2}+{\int }_a^b{f}_2(x{)}^2\right)}^{1/2}\\ & \le {\left({\int }_a^b{f}_1^2(x)\right)}^{1/2}+{\left({\int }_a^b{f}_2^2(x)\right)}^{1/2}\end{array}$$

Theorem: The Discrete Metric

Let $X$ be any set. For any two points $p,q\in X$, we have metric known as discrete metric

$$d(p,q)=\{\begin{array}{ll}0& p=q\\ 1& p\ne q\end{array}$$

Generalized Metric Space Definitions §

With metric spaces, we can generalize many of the definitions we had previously, by using our metric as a “distance” function! In fact, many of our definitions (ex. open sets) were given in terms of the norm $||\cdot ||$.

Let $X$ be a metric space.

A sequence $\{p_k\}\subseteq X$ is said to converge to a point $p\in X$, if $\forall ϵ>0$, $\exists N\in \mathbb{N}$ such that

$$d(p_k,p)<ϵ\forall k\ge N$$

We call $p$ the limit of the sequence $\{p_k\}$. Note that by this definition, we can see that a sequence converges if and only if the real sequence $\{d(p_k,p)\}\to 0$.

We also have the following set definitions:

• For a point $p\in X$, $r>0$, the set
  $$B_r(p)=\{q\in X:d(q,p)<r\}$$
  is the open ball around $p$ in $X$.
• $A\subseteq X$ is open if $\forall p\in X$, $\exists r>0$ such that $B_r(p)\subseteq X$.
• $A\subseteq X$ is closed if $\forall \{p_k\}\subseteq A$, if $\{p_k\}\to p\in X$, then $p\in A$.
• For $A\subseteq X$, a point $p\in A$ is an interior point if $\exists r>0$ where $B_r(p)\subseteq A$.

Theorem: The Complementing Characterization

Let $X$ be a metric space, and $A\subseteq X$. Then, $A$ is open if and only if $A^c$ is closed in $X$.

Theorem: Open and Closure of $X$ in Itself

Let $X$ be a metric space. Then, $X$ is open in $X$, and $X$ is also closed in $X$!

Theorem: Intersection and Union of Closed / Open Sets

Let $X$ be a metric space.

Then,

• The union of a (potentially infinite) collection of open subsets of $X$ is open.
• The intersection of a finite collection of open subsets of $X$ is open.

Also,

• The union of a finite collection of closed subsets of $X$ is closed.
• The intersection of a (potentially infinite) collection of closed subsets of $X$ is closed.

Completeness and the Contraction Mapping Principle §

Let $X$ be a metric space. Then, a sequence $\{p_k\}\subseteq X$ is Cauchy if $\forall ϵ>0$, $\exists K$ such that

$$d(p_k,p_l)<ϵ\forall k,l\ge K$$

Theorem: Convergence and Cauchy Sequences

Every convergent sequence is a Cauchy sequence.

Example: Cauchy Sequences in $C([a,b],\mathbb{R})$

Consider a sequence $\{f_k\}\subseteq C([a,b],\mathbb{R})$. Then, from definition of the metric, $\{f_k\}$ is a Cauchy sequence if and only if $\forall ϵ>0$, $\exists N\in \mathbb{N}$ such that

$$|f_k(x)-f_l(x)|<ϵ\forall x\in [a,b]k,l\ge N$$

Note that this must hold, as the metric takes the maximum difference between the functions.

In other words, $\{f_k\}$ is cauchy if and only if it is uniformly cauchy (converges uniformly).

We say metric space $X$ is complete, if $\forall \{p_k\}$ Cauchy sequence in $X$, $\exists p\in X$ such that $\{p_k\}\to p$. In other words, all Cauchy sequences converge to a point in $X$.

Remark

The metric spaces $\mathbb{R},{\mathbb{R}}^n,C([a,b],\mathbb{R})$ are complete.

Example: Incomplete Space

We can find that $C([a,b],\mathbb{R})$ and

$$d_1(f,g)={\int }_a^b|f(x)-g(x)|dx$$

is not complete.

We form a series of functions that converge to a function that is not continuous (and thus not in our space).

Lets consider $f_n:[0,2]\to \mathbb{R}$ where

$$f_n(x)=\{\begin{array}{ll}{x}^n& 0\le x\le 1\\ 1& 1\le x\le 2\end{array}$$

Then,

$${\int }_0^2|f_n(x)-f(x)|dx=0$$

If $f(x)=0$ for $0\le x<1$, $f(x)=1$ for $1\le x\le 2$.

So, $f_n\to f$ with respect to the $d_1$ metric, so $f_n$ is Cauchy. However, there does not exist a $g$ that is continuous from $[0,2]$ with real values such that

$${\int }_0^2|f-g|dx=0$$

Theorem: Complete Subspaces

Let $X$ be a complete metric space, and $Y$ a subspace of $X$. Then, $Y$ is a complete metric space if and only if $Y$ is a closed subset of $X$.

A corollary from this is that, every closed subset of $R,{\mathbb{R}}^n,C([a,b],\mathbb{R})$ is a complete metric space (from the previous remark).

Let $X$ be a metric space, and $T:X\to X$ be a function. Then, $T$ is Lipschitz if $\exists M\ge 0$ such that

$$d(T(p),T(q))\le Md(p,q)$$

For all $p,q\in X$. In other words, the change in distance between any two points $p,q$ after the mapping is bounded by some value.

Note that the same $M$ must work for all $p,q$ pairs.

A Lipschitz function $T:X\to X$ is a contraction if $M<1$. That is, there exists a $0\le c<1$ such that

$$d(T(p),T(q))\le c\cdot d(p,q)$$

Lipschitz Functions on $\mathbb{R}$

If $f:\mathbb{R}\to \mathbb{R}$ differentiable, then $f$ is Lipschitz with constant $M$ if and only if $|f^{\prime }(x)|\le M$.

Proof

Proof $\to$ §

If $f$ is differentiable and Lipschitz, then by definition of Lipschitz functions, $\forall x,y\in \mathbb{R}$,

$$|f(x)-f(y)|\le M|x-y|$$

And

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

So, applying our Lipschitz definition,

$$|f^{\prime }(x)|\le \underset{h\to 0}{\lim }\left|\frac{f(x+h)-f(x)}{h}\right|\le M$$

Proof $\leftarrow$ §

Conversely, if $\forall x\in \mathbb{R}$, $|f^{\prime }(x)|\le M$, then we want to show that $\forall x,y$,

$$|f(x)-f(y)|\le M|x-y|$$

As $f$ is differentiable, by the mean value theorem, there exists a $c$ between $x,y$ such that

$$|f^{\prime }(c)|=\left|\frac{f(x)-f(y)}{x-y}\right|\le M$$

We multiply $x-y$ on either side to find our result.

If $X$ is a metric space, and $T:X\to X$ is a function, then $p\in X$ is called a fixed point if

$$T(p)=p$$

In other words, the point after being mapped does not change.

With some assumptions on $T$, we can guarantee the existence of a fixed point for a function $T$.

Theorem: The Contraction Mapping Principle

Let $X$ be a complete metric space, and suppose $T:X\to X$ is a contraction. Then, $T:X\to X$ has exactly one fixed point.

Proof

Proof (Uniqueness) §

Say $T(p_1)=p_1$, and $T(p_2)=p_2$. By contraction, we find a $0\le c<1$ such that

$$|T(p_1)-T(p_2)|\le c|p_1-p_2|⟹|p_1-p_2|\le c|p_1-p_2|$$

This is only possible if $p_1=p_2$.

Proof (Existence) §

Start with any $p_0\in X$. Let $p_1=T(p_0),p_2=T(p_1),\dots$. We show that the sequence of points $\{p_k\}$ is Cauchy, thus there exists a $p$ where $\{p_k\}\to p$, and $T(p)=p$.

First, look at the distance between adjacent terms in the sequence.

$$\begin{array}{rl}& d(p_2,p_1)=d(T(p_1),T(p_0))\le Cd(p_1,p_0)\\ & d(p_3,p_2)=d(T(p_2),T(p_1))\le Cd(p_2,p_1)\le C^{2}d(p_1,p_0)\\ & ⋮\\ & d(p_{k+1},p_k)\le C^{k}d(p_1,p_0)\end{array}$$

Thus, we find that

$$\begin{array}{rl}d(p_{k+l},p_k)& \le d(p_{k+l},p_{k+l-1})+d(p_{k+l-1},p)\\ & \le d(p_{k+l},p_{k+l-1})+d(p_{k+l-1},p_{k+l-2})+d(p_{k+l-1},p)\\ & \le \dots \\ & \le (C^{k+l-1}+C^{k+l-2}+\cdots +C^k)d(p_1,p_0)\\ & \le C^k\left(\sum _{m=0}^{\infty }{C}^m\right)d(p_1,p_0)\\ & =C^k\frac{1}{1-C}\end{array}$$

Thus, our sequence is Cauchy!

Bringing it Together §

Let $\{p_k\}\to p$. We show that $T(p)=p$.

We essentially show that $T$ is continuous, so that if $\{p_k\}\to p$, then $\{T(p_k)\}\to T(p)$.

$$d(T(p_k),T(p))\le Cd(p_k,p)\to 0$$

By squeeze. So, $\{T(p_k)\}\to T(p)$.

We know that $\{p_k\}\to p$, and $\{p_{k+1}\}\to T(p)$, so $T(p)=p$ by the uniqueness of limits.

Note that by this, Lipschitz functions are essentially a stronger form of continuity!

Example: The Contraction Mapping Principle

Let $X=\{F\in C([0,\frac{1}{2}],\mathbb{R}):0\le f(x)\le 4\forall x\in [0,\frac{1}{2}]\}$. Let $T:X\to C([0,\frac{1}{2}],\mathbb{R})$ be given as

$$T(f(x))=1+{\int }_0^{x}f(t)dt$$

Part 1 §

Show that $T$ maps $X$ onto $X$.

Take any $f\in X$. We wish to show that $T(f)\in X$.

$$\begin{array}{rl}T(f)& =1+{\int }_0^{x}f(t)dt\\ & \le 1+\frac{1}{2}\underset{x\in [0,1/2]}{\max }f(x)\\ & \le 1+\frac{1}{2}4\\ & \le 3\end{array}$$

And furthermore, because $f(x)\ge 0$,

$$T(f)=1+{\int }_0^{x}f(t)dt\ge 1$$

So, $1\le T(f)\le 3$, meaning $T(f)\in X$!

Part 2 §

Is $T:X\to X$ a contraction?

Take any $f_1,f_2\in X$. To show that $T$ is a contraction, we wish to show that for some $M<1$,

$$\begin{array}{rl}d(T(f_1),T(f_2))& \le Md(f_1,f_2)\\ \underset{x\in [0,\frac{1}{2}]}{\max }\left|{\int }_0^x{f}_1(t)-f_2(t)dt\right|& \le \underset{x\in [0,\frac{1}{2}]}{\max }{\int }_0^x|f_1(t)-f_2(t)|dt\\ & \le \frac{1}{2}\underset{x\in [0,\frac{1}{2}]}{\max }|f_1(t)-f_2(t)|\end{array}$$

We find $M=1/2$! So, $T$ is a contraction.

Part 3 §

Write down a fixed point of $T$.

For a fixed point of $T$,

$$T(f)=f⟹f(x)=1+{\int }_0^{x}f(t)dt$$

This gives us system

$$\{\begin{array}{l}{f}^{\prime }(x)=f(x)\\ f(0)=1\end{array}$$

Which has a solution of $e^x$. $e^x$ is a fixed point of our system!

This theorem is quite important in differential equations!

The Existence Theorem for Nonlinear Differential Equations §

Let $I$ be an open interval of real numbers, $x_0\in I$. Then, for $y_0$ and a function $h:I\to \mathbb{R}$, consider the differential system.

$$\{\begin{array}{ll}{f}^{\prime }(x)=h(x)& \forall x\in I\\ f(x_0)=y_0& \end{array}$$

We ask, does there exist a differentiable function $f:I\to \mathbb{R}$ that satisfies the above differential equation?

Previously, we established that given $h$ continuous, there does exist an $f$, and this $f$ is unique with formula

$$f(x)=y_0+{\int }_{x_0}^{x}h(t)dt\forall x\in I$$

Now, we consider much more general differential systems.

Suppose $O$ is an open subset of ${\mathbb{R}}^2$, and let $(x_0,y_0)\in O$. Also suppose that we have a continuous function $g:O\to \mathbb{R}$.

Let $I$ be an interval in $\mathbb{R}$, and define the function $f:I\to \mathbb{R}$ such that the set of inputs/outputs is contained within $O$.

$$\{(x,f(x)):x\in I\}\subseteq O$$

We ask, can we find this interval $I$ containing $x_0$, and a differentiable function $f(x)$ satisfying

$$\{\begin{array}{l}{f}^{\prime }(x)=g(x,f(x))\\ f(x_0)=y_0\end{array}$$

Note how now, the derivative of our $f$ is given in terms of $f$ as well!

We use this setup from here on.

Lemma: The Equivalence Lemma

Let $O$ be an open subset of ${\mathbb{R}}^2$ with the point $(x_0,y_0)$, and suppose we have $g:O\to \mathbb{R}$ continuous.

If $I$ is a neighborhood of the point $x_0$, and $f:I\to \mathbb{R}$ has the property that $(x,f(x))\in O$ for all $x\in I$, then the following two are equivalent:

• The function $f:I\to \mathbb{R}$ is differentiable and is a solution of the differential equation

  $$\{\begin{array}{ll}{f}^{\prime }(x)=g(x,f(x))& \forall x\in I\\ f(x_0)=y_0& \end{array}$$

• The function $f:I\to \mathbb{R}$ is continuous and is a solution of the integral equation

  $$f(x)=y_0+{\int }_{x_0}^{x}g(t,f(t))dt\forall x\in I$$

Proof

Proof $\to$ §

Assume (1). Then, $f$ differentiable implies $f$ continuous, and because $f^{\prime }(x)=g(x,f(x))$ (a composition of continuous functions), $f^{\prime }$ is also continuous. So, by the fundamental theorem of calculus, we find

$$\begin{array}{r}f(x)-y_0=f(x)-f(x_0)={\int }_{x_0}^x{f}^{\prime }(t)dt={\int }_{x_0}^{x}g(t,f(t))dt\end{array}$$

Which is formula 2.

Proof $\leftarrow$ §

Assume (2). Since $f(x)=y_0+{\int }_{x_0}^{x}g(t,f(t))dt$, and $g$ is continuous (as remarked earlier), we know the integral of a continuous function is differentiable. So, $f$ is differentiable, and

$$f^{\prime }(x)=g(x,f(x))$$

Also, $f(x_0)=y_0$ can be obtained as ${\int }_{x_0}^{x_0}g(t,f(t))dt=0$.

Let’s define a function $T(f)$ as the equation in point 2 of the equivalence lemma.

$$T(f)=y_0+{\int }_{x_0}^{x}g(t,f(t))dt$$

Using this function $T(f)$, with an additional constraint on $g$, we can guarantee the existence of a unique solution for our system! This is known as the Existence and Uniqueness Theorem.

Theorem: Existence and Uniqueness

Let $O$ be an open subset of the plane ${\mathbb{R}}^2$ that contains the point $(x_0,y_0)$. Suppose that the function $g:O\to {\mathbb{R}}^2$ is continuous.

Now, assume an additional constraint, that $\exists M>0$ such that $\forall (x,y_1),(x,y_2)\in O$,

$$|g(x,y_1)-g(x,y_2)|\le M|y_1-y_2|$$

Then, there is an open interval $I$ with $x_0$ such that the differential equation

$$\{\begin{array}{ll}{f}^{\prime }(x)=g(x,f(x))& x\in I\\ f(x_0)=y_0& \end{array}$$

Has exactly one solution.

Proof: Existence and Uniqueness

Because $O$ is open, we can choose numbers $a,b>0$ such that the box of width $2a$, height $2b$ centered around $(x_0,y_0)$ is contained within $O$.

$$R=[x_0-a,x_0+a]\times [y_0-b,y_0+b]\subseteq O$$

For some positive number $0<l\le a$, let $I_l$ be the closed interval $[x_0-l,x_0+l]$, and define

$$X_l=\{f:[x_0-l,x_0+l]\to [y_0-b,y_0+b],\text{Continuous}\}$$

Or in other words, the set of all continuous functions along our interval $I_l$. Note that the domain and image of these functions are contained within our box $R$ by assumption.

For a function $f\in X_l$, we define the function $T(f)$ in $C(I_l,\mathbb{R}$ as

$$T(f)(x)=y_0+{\int }_{x_0}^{x}g(t,f(t))dt\forall x\in I_l$$

We now prove two conditions:

1. $T$ maps the space $X_l$ to itself
2. $T:X_l\to X_l$ is a contraction

Then, by the Contraction Mapping Principle, we have a unique fixed point.

$T:X_l\to X_l$ §

Because we $g(x,y)$ is continuous on a compact set, we know that $\exists K$ such that $|g(x,y)|\le K$ for all $(x,y)$ in our box $R$.

Then, if $f\in X_l$, and $x\in I_l$ (so $x\in [x_0-l,x_0+l]$), we have

$$\begin{array}{rlr}|T(f)(x)-y_0|& =\left|{\int }_{x_0}^{x}g(t,f(t))dt\right|& \\ & \le |x-x_0|\underset{t\in [x_0-l,x_0+l]}{\max }|g(t,f(t))|& \\ & \le |x-x_0|K& \text{Bounds}\\ & \le l\ast K& \text{Domain of Function}\end{array}$$

So, if $Kl\le b$, the range of our function is bounded between $[y_0-b,y_0+b]$, meaning $T:X_l\to X_l$. We choose $l$ small enough to make this happen.

$T$ is a Contraction §

We now show that $T:X_l\to X_l$ is a contraction provided $l$ is sufficiently small. We want

$$d(T(f_1),T(f_2))\le Cd(f_1,d_2)$$

Where $d$ is the uniform metric.

$$\begin{array}{rlr}T(f_1)(x)-T(f_2)(x)& ={\int }_{x_0}^{x}g(t,f_1(t))dt-g(t,f_2(t))dt& \\ |T(f_1)(x)-T(f_2)(x)|& =\left|{\int }_{x_0}^{x}g(t,f_1(t))dt-g(t,f_2(t))dt\right|& \\ & \le {\int }_{x_0}^x|g(t,f_1(t))dt-g(t,f_2(t))|dt& \text{Integral Property}\\ & \le {\int }_{x_0}^{x}M(f_1(t)-f_2(t))dt& \text{Assumption on g}\\ & \le M(x-x_0)\underset{t\in [x_0-l,x_0+l]}{\max }|f_1(t)-f_2(t)|& \\ & \le M\cdot l\cdot \underset{t\in [x_0-l,x_0+l]}{\max }|f_1(t)-f_2(t)|=M\cdot l\cdot d(f_1,f_2)& \end{array}$$

So, we found that

$$d(T(f_1),T(f_2))\le Mld(f_1,f_2)$$

As $M$ is fixed, and we can determine $l>0$, we make $l$ as small as we want! So, provided that $Ml<1$, we have a contraction.

Thus, by the Contraction Mapping Principle, there is exactly one fixed point for $T$. In other words, there exists a unique $f:(x_0-l,x_0+l)\to (y_0-b,y_0+b)$ such that $f^{\prime }(x)=g(x,f(x)),f(x_0)=y_0$.

This is a really important proof!

Note that there are many constraints that this theorem assumes in order for our result to hold; below, we address some common misconceptions.

Remark: Failure of Uniqueness

There are examples of continuous $g$ for which uniqueness fails. Consider

$$f^{\prime }(t)=3f(t{)}^{2/3}f(0)=0$$

Then, we can find the following solutions that are non-unique.

$$f(t)=\{\begin{array}{ll}0& t\le 0\\ t^3& t\ge 0\end{array}f(t)=0$$

In fact, we can find any solution of the form

$$f(t)=\{\begin{array}{ll}0& t\le C\\ (t-C{)}^3& t\ge C\end{array}$$

And find that in fact, our initial problem fails the Lipschitz condition, as for $g(t,y)=3y^{2/3}$, $g_y$ is unbounded as $y\to 0$.

Remark: Existence is a Local Property

Consider system

$$f^{\prime }(t)=f^2(t)f(0)=0$$

Note that one solution to this is $f(t)=1/(1-t)$, which only works for $(-\infty <t<1)$.

Remark 3

Let $g:{\mathbb{R}}^2\to \mathbb{R}$, and assume that

$$|g(t,y_1)-g(t,y_2)|\le M|y_1-y_2|$$

For some $M>0$, and for all $(x,y_1),(x,y_2)$.

If $f_1^{\prime }(t)=g(t,f_1(t))$, $f_1(t_0)=y_0$, and $f_2^{\prime }(t)=g(t,f_2(t))$, $f_2(t_0)=y_0$, then $f_1(t)=f_2(t)$ for all $t$.

Proof

Look at $E(t)=(f_1(t)-f_2(t){)}^2$. We find that

$$\begin{array}{rl}{E}^{\prime }(t)& =2(f_1(t)-f_2(t))(f_1^{\prime }(t)-f_2^{\prime }(t))\\ & \le 2|f_1(t)-f_2(t)||g(t,f_1(t))-g(t,f_2(t))|\\ & \le 2M|f_1(t)-f_2(t)||f_1(t)-f_2(t)|=2ME(t)\end{array}$$

Thus, $E^{\prime }(t)\le 2ME(t)$, $E(t_0)=0$.

$$\begin{array}{r}{E}^{\prime }(t)-2ME(t)\le 0\\ \frac{d}{dt}E(t)e^{-2Mt}=E^{\prime }(t)e^{-2Mt}-2ME(t)e^{-2Mt}\le 0\end{array}$$

Because $E(t)e^{-2Mt}$ is monotonically decreasing, and it is 0 at $t_0$, and $E(t)e^{-2t}\ge 0$ for all $t$, we know that

$$E(t)=0$$

Which forces $f_1(t)=f_2(t)$.

Proof (2)

Let $f_1,f_2:\mathbb{R}\to \mathbb{R}$ differentiable, satisfying

$$f_1^{\prime }(t)=g(t,f_1(t))f_2^{\prime }(t)=g(t,f_2(t))$$

And $f_1(t_0)=f_2(t_0)=y_0$. Also assume the Lipschitz condition

$$|g(t,y_1)-g(t,y_0)|\le M|y_1-y_2|$$

From our main theorem, $\exists l>0$ such that $f_1(t)=f_2(t)$ $\forall |t-t_0|<l$.We wish to show that this holds for all $t$.

Look at the set $S=\{t:f_1(t)=f_2(t)\}$. We find that

• Because $f_1,f_2$ are continuous, then $S$ is closed!
• $S$ is open. Let $t_1\in S$. Thus, $f_1(t_1)=f_2(t_1)$, and there exists a $\delta >0$ such that $f_1(t)=f_2(t)$ $\forall |t-t_1|<\delta$, by from the local uniqueness theorem applied to $t_1$.

Because $S$ is open and closed, the only set that satisfies this is the empty set, or the entire space! Because $t_0\in S$, we find that $S$ must be the entire space.