tea.mathoverflow.net - Discussion Feed (Can a function have a derivative which is not ITSELF a function?)

AndrewL comments on "Can a function have a derivative which is not ITSELF a function?" (12290)

AndrewL — Wed, 29 Dec 2010 07:34:33 -0800

@ Pete Please don't get frustrated. You're right,equivalent in the mathematical sense means "iff" rather then "if",so again,my bad. It's a small linguistic error,but this is mathematics,so it means everything. From my context,it's clear I meant "implies".I'm still learning and I have a tendency to be careless in informal discussion forums like this. It's no excuse,though-these are professionals in here and I need to work to be more careful.

@Ryan Which is why I agreed to let it die.

Ryan Budney comments on "Can a function have a derivative which is not ITSELF a function?" (12279)

Ryan Budney — Wed, 29 Dec 2010 01:44:25 -0800

I agree with Alex and Mariano. This is a forum for talking about MO, not for solidifying people's understanding of undergraduate mathematics.

Pete L. Clark comments on "Can a function have a derivative which is not ITSELF a function?" (12278)

Pete L. Clark — Wed, 29 Dec 2010 01:40:57 -0800

@ AndrewL:

You wrote: "If you allow the function to be complex differentiable (which is equivalent to saying it's infinitely differentiable as a vector function of 2 real valued components)".

This is false. If you had said "implies" it would be true, but you said "equivalent".

Your writing consistently contains errors of spelling, punctuation, grammar, logic and mathematics. You create the impression of being unable or unwilling to get these details correct. For a while I thought I could be of some help with this, but at this point I realize I am quite frustrated. I have decided to stop responding to what you write, at least for now.

Mariano comments on "Can a function have a derivative which is not ITSELF a function?" (12277)

Mariano — Wed, 29 Dec 2010 00:49:36 -0800

It never hurts, sure. But the point Alex is making is that this is completely off-topic here.

AndrewL comments on "Can a function have a derivative which is not ITSELF a function?" (12276)

AndrewL — Wed, 29 Dec 2010 00:43:14 -0800

@Alex Never hurts to get feedback from professionals on the simple things. I won't get this quality and depth of feedback from undergraduates.It seems to me to be a question worth tossing around by professionals from time to time since a lot of us-especially when graduate students-are so busy trying to lay the foundations for research in our temples as quickly as possible,that we tend to accept a lot of things without questioning them. It's important to question things we take for granted once in awhile-if only to see if we can refine thier definitions in ways that advance research.

That being said-this one's pretty much exhausted, so I'll let it die now. Thanks for all the feedback!

Alex Bartel comments on "Can a function have a derivative which is not ITSELF a function?" (12274)

Alex Bartel — Wed, 29 Dec 2010 00:09:43 -0800

AndrewL, your original question was "is this undergraduate analysis question appropriate for MO?" and it received a pretty clear answer. Do you agree that the discussion of the actual mathematics doesn't belong on meta.MO either? You have been pointed to several sites, where this question and the ensuing discussion would be appropriate. So why not take it there? I find it strange that an undergrad analysis topic is discussed at such length on meta.MO. Maybe it's just me.

AndrewL comments on "Can a function have a derivative which is not ITSELF a function?" (12273)

AndrewL — Tue, 28 Dec 2010 23:10:09 -0800

@Pete You either misunderstood what I said or I'm misunderstanding you,not sure which yet: "@AndrewL: FYI, I see two problems with your last post.

(i): Your characterization of complex differentiable functions is not correct. For instance, the function f(x+iy) = x-iy is infinitely differentiable as a function from R^2 to R^2. You are missing the Cauchy-Riemann equations."

By a function being "infinitely differentiable as a vector function of 2 real valued components",I meant the function has continuous partial derivatives of all orders as a function from R^2 to R^2. This,of course,is equivelent to saying the function is infinitely differentiable in R^2 at the point in question (at least,if I remember my vector analysis in R^n correctly).Clearly,the "conjugation" function defined in your post also fits that criteria.

I'm well aware of the CR equations and the resulting relationship of complex differentiablity with infinite real differentiabilty at a point in the plane. Every complex differentiable function is infinitely differentable as a function from R^2 to R^2,BUT THE CONVERSE IS NOT TRUE BY THE CR EQUATIONS. In fact,the only way a real valued function can satisfy the CR equations is if it's constant! (This is a surprisingly subtle point that a lot of people forget after thier first time around in complex variables. I wish I had a dollar for every graduate student I've seen that blows this when asked off the top of thier head if they haven't reviewed thier basic complex function theory.)

(ii): The modulus of the rotation does not (necessarily) shrink as one approaches the limit: it approaches |f'(z_0)|.

Oopsie,you're right,my bad.

@Deane Complex differentiability is indeed an extraordinarily strong condition,which is why the classical theory of functions of a complex variable is considered somewhat limited unless one considers functions defined on regions with singularities.

And people ask me why I won't choose analysis as one of my qualifying exam topics.........LOL

deane.yang comments on "Can a function have a derivative which is not ITSELF a function?" (12272)

deane.yang — Tue, 28 Dec 2010 13:41:26 -0800

[deleted]

Pete L. Clark comments on "Can a function have a derivative which is not ITSELF a function?" (12248)

Pete L. Clark — Mon, 27 Dec 2010 05:17:39 -0800

[deleted upon request]

deane.yang comments on "Can a function have a derivative which is not ITSELF a function?" (12247)

deane.yang — Sun, 26 Dec 2010 21:34:37 -0800

And the discussion up to this point has been on functions from a domain in $R^n$ to $R^m$. Complex differentiability is, as Pete says, a stronger condition and implies much more.

deane.yang comments on "Can a function have a derivative which is not ITSELF a function?" (12246)

deane.yang — Sun, 26 Dec 2010 21:33:00 -0800

AndrewL, I haven't any idea what stage you are in your studies or research, but my advice is to keep things as simple as possible. Don't worry about hard stuff (and for me [unnecessary comment deleted] the statement "derivative might fail to exist at a countable number of points" counts as hard stuff) until you really need it for something. My statement was overly vague and meant to be a "visual" statement (that I would use, say, in non-rigorous calculus courses). It can be made precise in different ways. I suggest that unless you are doing something that requires working with functions that are not necessarily differentiable everywhere, focus on the situation where the function is smooth (infinitely differentiable everywhere).

Pete L. Clark comments on "Can a function have a derivative which is not ITSELF a function?" (12245)

Pete L. Clark — Sun, 26 Dec 2010 20:57:07 -0800

@AndrewL: FYI, I see two problems with your last post.

(ii): The modulus of the rotation does not (necessarily) shrink as one approaches the limit: it approaches |f'(z_0)|.

[(iii): redacted. It seemed unnecessary in retrospect.]

AndrewL comments on "Can a function have a derivative which is not ITSELF a function?" (12243)

AndrewL — Sun, 26 Dec 2010 19:05:27 -0800

@Deane I assume by "sufficiently smooth",you mean the derivative fails to exist AT AT MOST a countable number of points in it's domain. If you allow the function to be complex differentiable (which is equivalent to saying it's infinitely differentiable as a vector function of 2 real valued components), the derivative has an even more interesting geometric interpretation: Under the "microscope", it becomes a convergent sequence of rotations in the complex plane whose modulus shrinks as one approaches the limit!

deane.yang comments on "Can a function have a derivative which is not ITSELF a function?" (12242)

deane.yang — Sat, 25 Dec 2010 14:05:08 -0800

My view of the derivative (no matter what the dimensions of the domain and range are): Any sufficiently smooth function, when magnified enough, looks like a linear function.

Dylan Wilson comments on "Can a function have a derivative which is not ITSELF a function?" (12241)

Dylan Wilson — Sat, 25 Dec 2010 12:26:14 -0800

@Andrew- I'm also a bit confused by your distinction between derivatives for one-variable and derivatives in n-dimensions (n>1). Derivatives (at a point) in one variable are also linear maps... they are given by 1 x 1 matrices and are maps from R --> R. In particular, they are functions :)

If you're looking for a more chocolatey definition of the derivative that looks more like the higher dimensional ones, then how about:

The derivative of f at p is the unique linear map L: R ---> R such that f(p+h) = f(p) + L(h) + e(h), where e is some function satisfying e(h)/h ---> 0 as h ---> 0. Of course... this is precisely the same as the ``vanilla'' definition.

AndrewL comments on "Can a function have a derivative which is not ITSELF a function?" (12215)

AndrewL — Thu, 23 Dec 2010 18:52:11 -0800

@Todd Your interpretation is the correct one. I'm clearly overthinking this. One of the derivative's great virtues is that it makes precise the idea of analysis as reversible linearization of differentiable functions and does so in a very simple manner. It's clear the answer is no when looked at from that perspective.

Todd Trimble comments on "Can a function have a derivative which is not ITSELF a function?" (12204)

Todd Trimble — Thu, 23 Dec 2010 12:23:55 -0800

@gilkalai: I could be wrong, but when I read "I think I'm an idiot who overthinks the simple things,that's what", I didn't take that as sarcasm or hostility, but rather as AndrewL's berating himself over what he now regards as a simple question.

Kevin Lin comments on "Can a function have a derivative which is not ITSELF a function?" (12201)

Kevin Lin — Thu, 23 Dec 2010 11:08:44 -0800

I mean, it is easy to construct a function f(x) for which lim_{x -> 0^+} f'(x) and lim_{x -> 0^-} f'(x) both exist but don't agree. Is something like this what you're talking about?

In any case, I also agree that math.stackexchange would be a better venue for this question.

Alex Bartel comments on "Can a function have a derivative which is not ITSELF a function?" (12195)

Alex Bartel — Thu, 23 Dec 2010 07:57:18 -0800

Dear AndrewL, I would like to third (or fourth?) the recommendation that you use math.stackexchange.com for questions on undergraduate material. There is no reason for you to expect that you might receive a less qualified answer there than here, if indeed that is the reason for your reluctance. Also, if your question turns out to be not well-posed or misguided in some way, people will be much more patient and forgiving there than here.

Mariano comments on "Can a function have a derivative which is not ITSELF a function?" (12194)

Mariano — Thu, 23 Dec 2010 07:32:08 -0800

gil said:

It is not clear what you mean by "Start questioning all the little things you took for granted as an undergraduate".

...nor why such questioning should happen in MO :)

gilkalai comments on "Can a function have a derivative which is not ITSELF a function?" (12193)

gilkalai — Thu, 23 Dec 2010 07:24:12 -0800

Andrew, I think you still miss the point in Ryan's comment and mine. The relevant fact that Ryan mentioned to your question is that the limit is by definition a single point. So if the derivative is defined at a point it has a single value. Therefore by the ordinary definition if the derivative is defined it is a function.

"@gilkalai I think I'm an idiot who overthinks the simple things,that's what."

This is an unfair reaction. You asked our feedback about the question itself and about if it is appropriate to MO. The question to you is if you still think the question is appropriate to MO. I dont see any reason to be sarcastic or hostile about it. [It turned out that my reading of Andrew's reaction was incorrect. Thanks Todd and Andrew.]
In addition, your reaction makes it unclear if you agree that the question is not good and has no meaning as asked, and if you think it can be repaired so that it represents some nice question.

The distinction you make between n=1 and n>1 is still not clear.

"these are the kinds of dumb questions you ask as a graduate student when you start questioning all the little things you took for granted as an undergraduate"

We can discuss this particular issue further if you wish. Its ok to ask dumb question, but again there is another issue about asking them on MO. It is not clear what you mean by "Start questioning all the little things you took for granted as an undergraduate".

Anton Geraschenko comments on "Can a function have a derivative which is not ITSELF a function?" (12187)

Anton Geraschenko — Thu, 23 Dec 2010 04:40:48 -0800

I read the phrase "a derivative which is defined by a limit at a point but is not ITSELF a well defined function" as follows. The function $x \sin(1/x)$ has a well-defined derivative at every non-zero value of $x$, but at $x=0$, every slope between -1 and 1 is attainable as a limit. So you could define a "derivative" to be a subset of the plane which is the union of the graph of the derivative away from $x=0$ with the vertical segment {0}×[-1,1].

Supposing this is what you meant (or even if it's not), it's not clear to me what you're looking for in an answer.

AndrewL comments on "Can a function have a derivative which is not ITSELF a function?" (12185)

AndrewL — Thu, 23 Dec 2010 03:42:35 -0800

@gilkalai I think I'm an idiot who overthinks the simple things,that's what.

As for my distinction,I was referring to the total derivative in n-dimensional real Euclidean space,which can be defined by a linear transformation (which can be expressed in matrix form with a choice of basis). Where n=1,we get the usual plain vanilla derivative on the real line. So obviously from this context,the answer to my question is no and I feel like an idiot for asking. But these are the kinds of dumb questions you ask as a graduate student when you start questioning all the little things you took for granted as an undergraduate.

@Ryan Thank you for reminding me of a subtle and often forgotten fact from general topology. Clearly,every limit in the range of a real-valued function in R with the usual topology is a limit point in the topological sense,but the converse is not true. (My brain is too fried from insomnia to give a counterexample right now.)

gilkalai comments on "Can a function have a derivative which is not ITSELF a function?" (12182)

gilkalai — Wed, 22 Dec 2010 23:51:43 -0800

Andrew, to the question "Can a function have a derivative which is defined by a limit at a point but is not ITSELF a well defined function?" the answert is indeed NO. Since, as Ryan said, the limit is unique. So the question without you further thinking about it is not good for both sites. Do you agree?

If you are interested in generalized forms of derivatives i.e., which replace "limits" by something else then there are various generalized notions. Now that you look back on your question is this what you meant?

You also make a destinction between one dimension and many dimensions which I do not understand: "For derivatives in higher dimensional Euclidean spaces and manifolds, the answer is clearly no since the derivative is defined by a linear map in these cases."

So looking back on your question based on the remarks here, what do you think about it?

Pete L. Clark comments on "Can a function have a derivative which is not ITSELF a function?" (12181)

Pete L. Clark — Wed, 22 Dec 2010 23:23:39 -0800

As stated, the question does not make sense to me. But there is a certain amount of goodwill shown on math.SE to such questions: i.e., often people will work with the OP to craft a sensible question, if possible.

Ryan Budney comments on "Can a function have a derivative which is not ITSELF a function?" (12180)

Ryan Budney — Wed, 22 Dec 2010 23:09:31 -0800

While we're on the topic of generalized derivatives, there's a lovely and widely-applicable 1972 paper of Golomb and Tapia on generalized gradients in Banach spaces.

Andy Putman comments on "Can a function have a derivative which is not ITSELF a function?" (12179)

Andy Putman — Wed, 22 Dec 2010 22:50:23 -0800

@Andrew L : Questions about standard matters of undergraduate mathematics (like why a function can have at most one derivative at any given point) are not appropriate for MO.

Ryan Budney comments on "Can a function have a derivative which is not ITSELF a function?" (12176)

Ryan Budney — Wed, 22 Dec 2010 22:39:31 -0800

I'm also confused about your statement that limits do not have to be single-valued. It sounds like you're thinking about cluster points, not limits. Limits are always single-valued in Hausdorff spaces.

Ryan Budney comments on "Can a function have a derivative which is not ITSELF a function?" (12175)

Ryan Budney — Wed, 22 Dec 2010 22:30:37 -0800

Oh, I'm not sure I understand what you're asking about then. The derivative in the single-variable context is always a function. If $f$ has domain $A$, the derivative $f'$ has a domain $A'$ and $A'$ is a subset of $A$. $A'$ is precisely the points of $A$ where the limit expression in the definition of the derivative exists.

AndrewL comments on "Can a function have a derivative which is not ITSELF a function?" (12174)

AndrewL — Wed, 22 Dec 2010 22:22:09 -0800

@Ryan I'm aware of the generalizations,but my question was reexamining the "plain vanilla" definition of a derivative of a function from real analysis of one variable as a limit of "quotients". It's not clear to me THIS always has to be itself a well defined function.

It IS quite interesting you bring up the concept of subderivative,which I'm not that familiar with-it was mentioned in my real analysis course in passing and it came up again in my advanced differential equations course. This is clearly always a well defined function since it is defined by one sided limits and these can be made functions by suitable domain restrictions.

Ryan Budney comments on "Can a function have a derivative which is not ITSELF a function?" (12173)

Ryan Budney — Wed, 22 Dec 2010 22:13:58 -0800

Your question doesn't sounds appropriate for MO -- certainly it would be appropriate for math.stackexchange -- mainly because it isn't all that focused. But there are more pliable notions of derivatives, for example the "weak derivative", stochastic calculus and such. Depending on the university this could either be an upper-level undergraduate topic or introductory graduate level material.

http://en.wikipedia.org/wiki/Stochastic_calculus
http://en.wikipedia.org/wiki/Weak_derivative
http://en.wikipedia.org/wiki/Subderivative

AndrewL comments on "Can a function have a derivative which is not ITSELF a function?" (12172)

AndrewL — Wed, 22 Dec 2010 22:00:51 -0800

As you get more sophisticated in mathematics,you begin to question simple things you took for granted as an undergraduate,even in theoretical courses. I was thinking about the formal definition of a derivative in calculus in various contexts and I began to wonder: Can a function have a derivative which is defined by a limit at a point but is not ITSELF a well defined function? For derivatives in higher dimensional Euclidean spaces and manifolds, the answer is clearly no since the derivative is defined by a linear map in these cases. But on the real line, it's not clear to me that the definition excludes cases where the derivative is a limit and NOT a function since limits aren't required to be singled valued for given a specific epsilon and delta "tolerance" for open intervals of a point. Yes,limits of functions are required to be unique for a specific point,but unless the function is continuous,this doesn't insure the limit determines the value of a function. ( Yes,every function with a derivative is continuous,but unless the map is smooth,this does not require the derivative to be continuous. Otherwise,the point is moot,of course.) I may be overthinking this and completely missing the obvious-which is why I asked the question on meta first. But do I have a legitimate concern here?