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In a recent post I asked what are the ways a continuous function cannot be smooth. Apparently my question was not well posed, but since I am still thinking about these types of functions I will try one more time to ask my question in a proper way.

I define a "kink" in the graph of a function as an abrupt, discontinuous change in the first derivative. For example, the function f(x) = |x| has what I call a kink at x=0. I apologize if this is not totally rigorous, but I think the idea of a "kink" is clear enough that someone could make it rigorous with little effort.

Besides sin(1/x) and the Weirstrauss function, what are some examples of non-smooth continuous functions without kinks, either in the graph of the function or the graph of any of it's derivatives?

(i.e. I am not looking for any function for which the graph of one of the derivatives has a kink)

In case anyone is interested, I edited the linked post yesterday to respond to Qiaochu's comment.

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Weierstrass :-) –  Mariano Suárez-Alvarez Aug 24 '10 at 23:10
    
So without kinks => derivative is continuous? I take it your definition of non-smooth is derivative is discontinuous? So a non-smooth function without kinks does not exist. Please define what you consider a smooth function and if it takes little effort to make the definition of kink rigorous, I suggest you do it and edit that into the question. –  Aryabhata Aug 24 '10 at 23:21
    
I am just using the standard definition of non-smooth, which to my understanding is that a non smooth function has only finitely many derivatives; and I also demand the function be at least continuous. So the functions I want are in C^0 for sure, can be in any C^k, and are not in C^infinity. Consider the Weirstrauss function, which has no derivative but is continuous everywhere. This function is certainly not smooth, but you also can't say its derivative is discontinuous, because it has no derivative. I think the idea of a kink is pretty clear, what part of my def do you object to? –  Matt Calhoun Aug 25 '10 at 0:12
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What is an 'abrupt discontinuous change' for instance? Is the fact that the derivative of |x| does not even exist at zero covered by your definition of 'abrupt discontinuous change'? Sorry, it might be pretty clear to you, but it is just ambiguous and vague words to me. –  Aryabhata Aug 25 '10 at 0:36

3 Answers 3

up vote 4 down vote accepted

This has already been said before, but let me iterate this. The value of the derivative of a function at a point is evaluated by computing a limit, if it exists. The limit of a sequence may not exist if at least one of the following happens:

  1. the absolute value of some of the terms grows without bounds;

  2. you can extract two subsequences which both converge, but they converge to different values.

It is important to keep in mind that both behaviors may happen for the same limit, and that for 2 you might be able to extract countably many subsequences each converging to a different value (only "countably many" as I am talking now of a limit of a sequence). EDIT: in fact, you can have uncountably many different limits (see Jason's comment below).

The situation for limits of functions is not simpler than the above, since you can "embed" any sequence into the limit defining the derivative of a continuous function at a point. One of the good consequences of the fact that you are dealing with a limit of a continuous function is that if situation 2 occurs and the two subsequences you found have limits a<b, then also every intermediate value $c \in [a,b]$ is the limit of a subsequence. An elaboration of this is that the derivative of a once differentiable function satisfies the intermediate value property (Darboux's Theorem).

Thus, if you analyze the behavior of the values of the limit that defines a derivative at a point you may find that, as you zoom in, the values you obtain

a. also zoom in on a single real number - the derivative exists;

b. are always contained in some finite interval, no matter how close to the actual limiting point you get - e.g. the function $x^2 \sin(1/x)$ near 0;

c. are unbounded (possibly with alternating signs) - e.g. the function $\sqrt{x} \sin (1/x)$.

Note also that further pathologies include the fact that the behaviors as you approach the limit from below or from above may well be different. Even crazier, the "overall" interval you find in b, or an infinite interval in the case c, may contain inside it also countably many of other intervals nested more or less arbitrarily.

To conclude, derivatives of continuous functions are so immensely complicated that there is very little you can say about them!

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The symbol for "less than" gave me quite a hard time! I hope that this post is legible now... –  damiano Aug 25 '10 at 0:14
    
This is only tangentially related to your post, but if you, say, have a sequence which enumerates the rationals in the usual way, I think that there is a subsequence which converges to any previously assigned real number. In particular, you can have more than a countable number of different subsequences which converge to different limits. –  Jason DeVito Aug 25 '10 at 1:45
    
@Jason: you are absolutely right! I am not sure how I missed this: thanks for spotting it! –  damiano Aug 25 '10 at 1:50

One rigorous definition of a kink that I can think of is that the left and right derivatives exist and do not agree. In that case, plenty of functions "don't have kinks" at points where they are continuous but not differentiable. $\sin \frac{1}{x}$ and the Weierstrass function "don't have kinks" in the sense that neither the left nor the right derivative exist (at $0$ for the first function and everywhere for the second), and one can come up with arbitrarily badly behaved examples (so again, I do not see the point of this question) as explained for example in damiano's comment to your previous post. If you do not understand this comment, I can elaborate.

Please don't take this the wrong way, but the fact that you ask this question and put the Weierstrass function in as an example indicates to me that you don't understand the construction of the Weierstrass function. The point of the construction of the Weierstrass function is that although a uniformly convergent sequence of continuous functions is continuous, a uniformly convergent sequence of differentiable functions need not be differentiable. One can imitate this construction for an extremely wide class of functions (e.g. by replacing the sines with another periodic function) and get any kind of examples you want. So when you say "the Weierstrass function," do you literally mean "the Weierstrass function," or do you mean "any function constructed sort of like the Weierstrass function"? If the latter, you should be more precise about this.

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I always appreciate your comments Qiaochu and I respect your opinion a lot. I know there are plenty of things I don't properly understand, thanks for taking the time to explain it! –  Matt Calhoun Aug 25 '10 at 0:15

A real valued function is said to have a discontinuity of the first kind at a point if the left and right-hand limits at that point exist but are unequal. There are, however, only countably many such points. These points, applied to the derivative, are kind of like "kinks." A discontinuity of the second kind is not a "kink" since one of the sided limits doesn't exist. (I am attempting, possibly incorrectly, to make precise your terminology.)

The derivative of $x^2 \sin(1/x^2)$ at zero is zero, so the function is differentiable everywhere, yet the derivative is unbounded in any neighborhood of the origin, which is thus a discontinuity of the second kind. More generally, Volterra's function gives a differentiable function with a derivative with an uncountable number (even positive measure) of discontinuities of the second kind.

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