# Showing Every Polygonal function is Lipschitz

There is a problem in my (N.L. Carothers') analysis textbook that asks:

Prove every polygonal function is Lipschitz. Thus, the Lipschitz functions are dense in $$C[a,b]$$.

Is this question supposed to say: prove every polynomial function is Lipschitz? My reason for thinking this is because polygonal functions are just polynomials with degree $$\leq 1$$, and therefore, would be constant and hence Lipschitz. On the other hand, if the author meant to ask this claim for polynomials in $$C[a,b] : = \{f:[a,b] \subset \mathbb{R} \to \mathbb{R}: f \ is \ continuous\}$$ we would only be studying polynomials on a compact domain in $$\mathbb{R}$$. Which, in turn would mean the polynomials we are considering are bounded and hence continuous.

I have definitely been staring at this problem for far too long, so any insight will be useful.

No, "polygonal" isn't a typo. A polygonal function on $$[a, b]$$ is formed by splitting $$[a, b]$$ into a finite set of intervals and defining an affine function on each interval, such that the affine functions "join up" at the interval boundaries. In other words, a function that looks like this:

A polygonal function is definitely not just a polynomial of degree at most one, and certainly isn't necessarily constant.

A more formal definition could be this:

A function on $$[a, b]$$ is polygonal iff here exists a strictly increasing finite sequence $$a=x_1<... such that for all $$i$$, the function is affine on $$[x_i, x_{i+1}]$$, and the function is continuous on $$[a, b]$$.

To prove such a function is Lipschitz, consider the following intuitions.

1. Think of the function $$x(t)$$ as representing the position of something over time. To say $$x$$ is $$\lambda$$-Lipschitz is precisely to say that the average speed of $$x$$ over any time interval is at most $$\lambda$$ (convince yourself of this).

2. To say $$x$$ is an affine function is precisely to say the particle travels at constant speed. Apply this to show an affine function is Lipschitz (and work out the Lipschitz constant).

3. To say $$x$$ is polygonal is precisely to say that the particle starts out traveling at speed $$v_1$$, then changes speed instantaneously to $$v_2$$, and so on a finite number of times. Think about why this makes the function Lipschitz (hint: there is a maximum speed $$\max v_i$$ the particle is ever travelling at).

In more detail, here is one way to do it.

Let $$f$$ be polygonal and let $$x < y$$ be two points in $$[a, b]$$. Let $$\xi_1 < ... < \xi_n$$ be the points in $$[a, b]$$ where $$f$$ "changes direction" that are between $$x$$ and $$y$$. Note that $$f'$$ is constant on each interval $$[\xi_i,\xi_{i+1}]$$, say $$f'=\lambda_i$$ on that interval. Similarly, $$f'$$ is constant on $$[x, \xi_1]$$ and on $$[\xi_n, y]$$, let's say equal to $$\alpha$$ and $$\beta$$ respectively. We have:

$$f(y)=f(x)+\alpha|\xi_1-x|+\sum_{i=1}^{n-1} \lambda_i |\xi_{i+1}-\xi_i|+\beta|y-\xi_n|$$

You can use this to obtain a bound on $$|f(y)-f(x)|$$ using the triangular inequality which will give you that $$f$$ is Lipschitz with constant $$\max(\alpha, \beta, \lambda_1, ..., \lambda_{n-1})$$.

• This is very helpful. Since polygonal is not a typo, how is a polygonal function defined? I can't find any useful information anywhere to begin to start thinking about how to prove that they are Lipschitz. Nov 19, 2020 at 19:14
• Also, I don't want this come off as me asking you to answer my original question for me - just reaching out for hints or helpful information :-). Nov 19, 2020 at 19:49
• @TaylorRendon I edited some thoughts into the answer. Nov 19, 2020 at 22:41
• Is there a way to do this showing that these affine functions are all lipschitz and that for arbitrary $u,a$ in the polygonal function, one can combine them along the endpoints of the intervals using the lipschitz condition for each and some triangle inequality properties? It seems like there should be a way to get a constant since there are a finite number of constants since there are a finite number of affine intervals. Nov 19, 2020 at 22:52
• @TaylorRendon Sure. Nov 24, 2020 at 20:39