# The Unapologetic Mathematician

## Continuously Differentiable Functions are Locally Lipschitz

It turns out that our existence proof will actually hinge on our function satisfying a Lipschitz condition. So let’s show that we will have this property anyway.

More specifically, we are given a $C^1$ function $f:U\to\mathbb{R}^n$ defined on an open region $U\subseteq\mathbb{R}^n$. We want to show that around any point $p\in U$ we have some neighborhood $N$ where $f$ satisfies a Lipschitz condition. That is: for $x$ and $y$ in the neighborhood $N$, there is a constant $K$ and we have the inequality

$\displaystyle\lVert f(y)-f(x)\rVert\leq K\lVert y-x\rVert$

We don’t have to use the same $K$ for each neighborhood, but every point should have a neighborhood with some $K$.

Infinitesimally, this is obvious. The differential $df(p):\mathbb{R}^n\to\mathbb{R}^n$ is a linear transformation. Since it goes between finite-dimensional vector spaces it’s bounded, which means we have an inequality

$\displaystyle\lVert df(p)v\rVert\leq\lVert df(p)\rVert_\text{op}\lVert v\rVert$

where $\lVert df(p)\rVert_\text{op}$ is the operator norm of $df(p)$. What this lemma says is that if the function is $C^1$ we can make this work out over finite distances, not just for infinitesimal displacements.

So, given our point $p$ let $B_\epsilon$ be the closed ball of radius $\epsilon$ around $p$, and choose $\epsilon$ so small that $B_\epsilon$ is contained within $U$. Since the function $df(p)$ — which takes points to the space of linear operators — is continuous by our assumption, the function $p\mapsto\lVert df(p)\rVert_\text{op}$ is continuous as well. The extreme value theorem tells us that since $B_\epsilon$ is compact this continuous function must attain a maximum, which we call $K$.

The ball is also “convex”, meaning that given points $x$ and $y$ in the ball the whole segment $x+t(y-x)$ for $0\leq t\leq1$ is contained within the ball. We define a function $g(t)=f(x+t(y-x))$ and use the chain rule to calculate

$\displaystyle g'(t)=df(x+t(y-x))\frac{d}{dt}(x+t(y-x))=df(x+t(y-x))(y-x)$

Then we calculate

\displaystyle\begin{aligned}f(y)-f(x)&=g(1)-g(0)\\&=\int\limits_0^1g'(t)\,dt\\&=\int\limits_0^1df(x+t(y-x))(y-x)\,dt\end{aligned}

And from this we conclude

\displaystyle\begin{aligned}\lVert f(y)-f(x)\rVert&=\left\lVert\int\limits_0^1df(x+t(y-x))(y-x)\,dt\right\rVert\\&\leq\int\limits_0^1\lVert df(x+t(y-x))(y-x)\rVert\,dt\\&\leq\int\limits_0^1\lVert df(x+t(y-x))\rVert_\text{op}\lVert(y-x)\rVert\,dt\\&\leq\int\limits_0^1K\lVert(y-x)\rVert\,dt\\&=K\lVert(y-x)\rVert\end{aligned}

That is, the separation between the outputs is expressible as an integral, the integrand of which is bounded by our infinitesimal result above. Integrating up we get the bound we seek.

May 4, 2011

## The Existence and Uniqueness Theorem of Ordinary Differential Equations (statement)

I have to take a little detour for now to prove an important result: the existence and uniqueness theorem of ordinary differential equations. This is one of those hard analytic nubs that differential geometry takes as a building block, but it still needs to be proven once before we can get back away from this analysis.

Anyway, we consider a continuously differentiable function $F:U\to\mathbb{R}^n$ defined on an open region $U\subseteq\mathbb{R}^n$, and the initial value problem:

\displaystyle\begin{aligned}v'(t)&=F(v(t))\\v(0)&=a\end{aligned}

for some fixed initial value $a\in U$. I say that there is a unique solution to this problem, in the sense that there is some interval $(-a,a)$ and a unique function $v:(-a,a)\to\mathbb{R}^n$ satisfying both conditions.

In fact, more is true: the solution varies continuously with the starting point. That is, there is an interval $I$ around $0\in\mathbb{R}$, some neighborhood $W$ of $a$ and a continuously differentiable function $\psi:I\times W\to U$ called the “flow” of the system defined by the differential equation $v'=F(v)$, which satisfies the two conditions:

\displaystyle\begin{aligned}\frac{\partial}{\partial t}\psi(t,u)&=F(\psi(t,u))\\\psi(0,u)&=u\end{aligned}

Then for any $w\in W$ we can get a curve $v_w:I\to U$ defined by $v_w(t)=\psi(t,w)$. The two conditions on the flow then tell us that $v_w$ is a solution of the initial value problem with initial value $w$.

This will take us a short while, but then we can put it behind us and get back to differential geometry. Incidentally, the approach I will use generally follows that of Hirsch and Smale.

May 4, 2011