# Thread: Differentiable function

1. I came to think about a question which I cannot really figure out. Suppose that is a function continuous on some interval . Also suppose that exists for some . Would this imply that is also differentiable at some other point on the interval, that is, is it possible for a continuos function to be differentiable at only one point?  2.

3. Originally Posted by thyristor I came to think about a question which I cannot really figure out. Suppose that is a function continuous on some interval . Also suppose that exists for some . Would this imply that is also differentiable at some other point on the interval, that is, is it possible for a continuos function to be differentiable at only one point?
Yes, a common example of such a function can be found here.  4. But that function is only continuous at x=0, and in my question I stated that the function be continuous on some interval.  5. If it's continuous on the interval and differentiable for some value t on the interval where t is as you've stated then it's differentiable everywhere on that interval, endpoints excepted. The way you've defined t makes this so and the fact that it's continuous on the interval means that it has no points of discontinuity.  6. Your conclusion seems reasonable, but I do not quite see how you arrive at it.  7. At second thought, I actually want to withdraw my statement about the conlusion being reasonable. Consider for example , defined for . Obviously is continuous on any interval or , with . It is easily shown that is not differentiable at or at (if you doubt it, I'll show you). However, is differentiable for all other . These facts combined contradict your conlusion.  8. You're right. It has to be differentiable at every point in the interval. Continuity is insufficient. It would to be differentiable for every point t in the interval, not just for some point t. That's what I get for providing a quick answer!  9. Any other input on this issue?  10. If the derivative function f'(x) exists at one point in your interval, and is in itself continuous, then all other points in your interval will be differentiable as well. Is this what you were looking for ?

Reference : Differentiable function - Wikipedia, the free encyclopedia

On the other hand it is possible for a function to be everywhere continuous, but nowhere differentiable, like the Weierstrass function for example.  11. Originally Posted by Markus Hanke If the derivative function f'(x) exists at one point in your interval, and is in itself continuous
I don't quite see what you mean by this. If a function is only defined at one point in an interval, it has no meaning to say that the function, in this case f'(x), is continuous.

I do know about the Wierstrass function. Note that this is a function that is nowhere differentiablem. What I asked was, if there is such a function that is continuous on some interval [a,b], and differentiable for some k, satisfying a<k<b, yet not differentiable for any other x in [a,b].  12. So what you mean is whether there is a function that is, though continuous throughout a given interval, only differentiable in one point within that interval ?  13. Originally Posted by Markus Hanke So what you mean is whether there is a function that is, though continuous throughout a given interval, only differentiable in one point within that interval ?
Yes, that is what I mean.  14. I do not believe that this is possible. It would mean that the derivative of that function is only defined for a specific, single value k, and nowhere else, whereas the original function remains continuous everywhere. Now you can of course set out with a function that is only defined for a single given value, but if you reverse the operation and integrate that function you don't get anything that is continuous along an interval.I can't offer formal proof though at the moment.  f(x)=2 for x<=0
f(x)=Weierstrass(x) for x>0

Now consider the interval x=[0...1]. f(x) is continuous in the interval, but differentiable only at x=0.

Perhaps a silly example, but it does fulfil your requirements.  16. My apologies, I am just now noticing that, in your original post, you are excluding the endpoints of your interval. This of course invalidates my example.
In that case I still maintain that this is not possible, though I can't offer formal proof.  17. My apologies, I am just now noticing that, in your original post, you are&nbsp;<em>excluding</em> the endpoints of your interval. This of course invalidates my example.&nbsp;<br>In that case I still maintain that this is not possible, though I can't offer formal proof.  18. I am referring to page 3 of the following document :

http://www.math.wfu.edu/tutorials/Ma...ntiability.pdf

According to this, your function would need to fulfill the following criteria in order to be differentiable at only one point k :

1) The function is NOT continuous except at x=k OR
2) Every point on the interval x<>k is a corner point OR
3) The function is undefined for every x<>k

(1) and (3) violate the conditions set out in your original post, whereas (2) is impossible since a corner point requires differentiability of the function on both sides of the point to be defined.
It appears the answer is therefore that such a function as your stipulate in the original post cannot exist.  19. Hi!

Sorry for my late reply; I have been quite busy the last days and forgot about our ongoing discussion. I thank you for the document you provided, but actually I don't think your conclusion is quite correct. To demonstrate this, take for example the Weirstrass function. Just like my function, it is continous and defined at all points on some interval, so (1) and (3) from your last posts cannot hold. So in order for the Wierstrass function not to be differentiable, every point on some interval has to be a corner point. Following your own argument, we conclude that this implies the existence of RHD and LHD, violating the definition of the Weierstrass function and thus, incorrectly, proving it cannot exist.

The problem here is that corner-point either has to be redifined, or the statement in the provided document has to be redfined.  20. Originally Posted by thyristor Hi!

Sorry for my late reply; I have been quite busy the last days and forgot about our ongoing discussion. I thank you for the document you provided, but actually I don't think your conclusion is quite correct. To demonstrate this, take for example the Weirstrass function. Just like my function, it is continous and defined at all points on some interval, so (1) and (3) from your last posts cannot hold. So in order for the Wierstrass function not to be differentiable, every point on some interval has to be a corner point. Following your own argument, we conclude that this implies the existence of RHD and LHD, violating the definition of the Weierstrass function and thus, incorrectly, proving it cannot exist.

The problem here is that corner-point either has to be redifined, or the statement in the provided document has to be redfined.
You are actually right, well spotted !
In that case I am more or less out of ideas, because I just cannot think of any function that would fit your requirements. At the same time I don't know how to formally prove that it cannot exist.

This is actually a really interesting problem. I will think about this a little more...  21. Neither can I think of any function satisfying my requirements, but I am unable to prove that such a function cannot exist. Let's hope someone else can shed some light on this matter, or that we solve it for ourselves first.  Bookmarks
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