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I am interested in continuous functions which are not smooth, i.e. which have only finitely many derivatives (for now just taking the case $f:\mathbb{R}\rightarrow\mathbb{R}$). As far as I can tell, there are only 2 features or properties of such a function which can make it non-smooth.

  1. The derivative of some order diverges to infinity at one or more points.

  2. The derivative of some order oscillates like $\sin(1/x)$ does at $x = 0$ at one or more points.

Q1: Are there any other ways for a continuous function to be non-smooth?

Q2: What are some example's of non-smooth functions which satisfy a differential equation?


In response to Qiaochu comment:

A simple "description of a limit failing to exist which is better than "the limit fails to exist" is to specify, for example, if the limit of a sequence of natural numbers diverges to infinity or not. for example: $1,2,3,...$ and $1,2,1,2,1,...$

These two examples are similar to my 1. and 2., i.e. the derivative diverges to infinity or it oscillates around without settling down.

I don't see the problem with trying to classify the ways continuous functions cannot be smooth by looking at their graphs. One example of a non smooth function is $\sin(1/x)$, it has a certain property at $x=0$, another non-smooth function is $|x|$ and it has another property at $x=0$.

I understand you feel the question is not well defined because "being like $\sin(1/x)$" is an intuitive concept.

Ok, so what are examples of non-smooth continuous functions without "kinks" like $\sin(1/x)$ has at $0$?

There are two that I can think of, $\sin(1/x)$ and the Weierstrass function. But if you think about the self similarity of the graph of the Weirstrauss function, then it's clear that the derivative of this function osscilates rapidly at every point like $\sin(1/x)$ does at $x=0$.

Regardless of whether or not it's well-defined to say the Weierstrass function is like $\sin(1/x)$ in some sense, I am just curious to know of some other examples of non-smooth continuous functions without kinks in their graph, can someone please help?

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The absolute value function seems to be not covered in your "list". –  damiano Aug 23 '10 at 16:27
I don't know that this is a well-defined question. It is not clear to me what the second example means rigorously, and I don't know what kind of answers you are expecting. A derivative fails to exist when a certain limit fails to exist, and I don't know of any description of a limit failing to exist which is better than "the limit fails to exist." –  Qiaochu Yuan Aug 23 '10 at 17:08
To emphasize Qiaochu's comment: given any sequence of real numbers $(a_n)$, you can find a continuous function $f \colon \mathbb{R} \to \mathbb{R}$ and a decreasing sequence $(t_n)$ of positive real numbers such that for every $n$ we have $\frac{f(t_n)-f(0)}{t_n} = a_n$. In particular, the limit defining the derivative may be "as bad as you want it to be". You can even make sure that $f$ is smooth away from the origin. –  damiano Aug 23 '10 at 17:14
How about fractals (eg Weiner process) which are nowhere smooth? –  KalEl Aug 23 '10 at 19:18
@Qiaochu: See the edit to my OP for my response. –  Matt Calhoun Aug 24 '10 at 4:39

4 Answers 4

To answer your second question, I offer two classical examples:

1: The cycloid


2: The tractrix


($a$ is a parameter in both examples)

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For Q1)

Depends on your definition of oscillate I suppose.

$f(x) = |x|$ does not have a derivative at $0$ and would not fit your criteria for 1 or 2.

In any case, your question seems to be equivalent to the question:

"In what ways can the limit of a function fail to exist at a point".

As to the comment of derivative of |x| being a step function and so the second derivative of |x| at 0 is infinity is nonsense, IMO. The first derivative itself does not exist, so talking about the second makes no sense. Perhaps you need to clarify exactly what you mean.

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The graph of the derivative of f(x)=|x| is a step function, so in this case the slope of the the tangent line at x=0 is infinite, i.e. this function is not smooth because the derivative diverges to infinity at x=0. –  Matt Calhoun Aug 23 '10 at 16:42
@Matt: f(x) = |x| is not a step function! In fact it is continuous everywhere. –  Aryabhata Aug 23 '10 at 16:43
Matt said the derivative of |x| (which is sgn x) is a step function. –  J. M. Aug 23 '10 at 16:45
@J. Mangaldan: Ah I see what he is saying. I misread. The derivative (of |x|) does not exist of 0, so talking about the derivative of the derivative of |x| at 0 does not make any sense to me! Matt, perhaps you should clearly define what you really mean. –  Aryabhata Aug 23 '10 at 16:47
Indeed, a poor choice of words. Matt, there is such a thing as "not having a derivative" at a point. Have you encountered the concept of "directional derivatives" in your studies? –  J. M. Aug 23 '10 at 16:55

Re Q2: For smooth ODEs (i.e. something like y′=f(y) for f smooth, possibly with y taking values in Rn but depending on one parameter), existence of a solution implies regularity (i.e. smoothness) of it. This isn't particularly easy to prove, though; I know a proof is in Lang's Differential and Riemannian Manifolds, and if I recall correctly the idea is to prove inductively that $y'$ is $C^1$ (or at least Lipschitz) by some sort of Gronwall inequality argument, and then get that $y'$ satisfies its own differential equation, and repeat inductively. The details are very hazy in my mind.

This is part of the general theory of ordinary differential equations, though, so there are probably plenty of sources.

For PDEs, automatic regularity is not necessarily the case, though it is for elliptic PDEs with smooth coefficients. More generally, regularity holds for certain hypoelliptic operators (e.g. the heat operator).

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One thing you might be interested in is Darboux's Theorem, which tells you that derivatives have the intermediate value property. There's a decent Wikipedia page on the topic and there was also a nice proof given by Sam B. Nadler, Jr. in the Math Monthly a little while back (Feb. 2010).

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