The Unapologetic Mathematician

Mathematics for the interested outsider

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.

About these ads

May 4, 2011 - Posted by | Analysis, Differential Equations

3 Comments »

  1. [...] on . We pick so that satisfies a Lipschitz condition on , which we know we can do because is locally Lipschitz. Since this is a closed ball and is continuous, we can find an upper bound for . Finally, we can [...]

    Pingback by The Picard Iteration « The Unapologetic Mathematician | May 5, 2011 | Reply

  2. [...] our proof that is locally Lipschitz involved showing that there’s a neighborhood of where we can bound by . Again we can pick a [...]

    Pingback by Another Existence Proof « The Unapologetic Mathematician | May 10, 2011 | Reply

  3. sure

    Comment by pasbu | October 16, 2012 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 366 other followers

%d bloggers like this: