# 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 - Posted by | Analysis, Differential Equations Comment by pasbu | October 16, 2012 | Reply