To understand the Darboux theorem

  • A
  • Thread starter feynman1
  • Start date
  • Tags
    Theorem
In summary: No, I haven't looked at it specifically. I'm more interested in understanding why the derivative is discontinuous at x=0.Have you looked at the derivative of this function near 0? Draw a graph, and evaluate the derivative at 0...No, I haven't looked at it specifically. I'm more interested in understanding why the derivative is discontinuous at x=0.
  • #1
feynman1
435
29
Darboux theorem says a derivative function must have an intermediate value theorem without requiring the derivative function to be continuous. Why is this property not true for any continuous function in its intermediate value theorem?
 
Physics news on Phys.org
  • #2
feynman1 said:
Darboux theorem says a derivative function must have an intermediate value theorem without requiring the derivative function to be continuous. Why is this property not true for any continuous function in its intermediate value theorem?
What do you mean? If we have any continuous function, then we do not have necessarily a derivative. So how could we make a statement about derivatives?
 
  • #3
This is to compare intermediate value thm and Darboux thm. Clear?
 
  • #4
feynman1 said:
This is to compare intermediate value thm and Darboux thm. Clear?
Not at all. You asked:
feynman1 said:
Why is this property ...
[Darboux: ##f'## takes every value between ##f'(a)## and ##f'(b)##]
feynman1 said:
... not true for any continuous function in its intermediate value theorem?
Continuous functions in general do not have derivatives. Hence the property cannot hold since the statement is void.

Did you mean why is it possible that ##f'## isn't continuous?
 
  • Like
Likes feynman1
  • #5
No. Why does a derivative function possesses a property of taking any intermediate value without declaring the derivative is continuous?
 
  • #6
feynman1 said:
No. Why does a derivative function possesses a property of taking any intermediate value without declaring the derivative is continuous?
The reason is, that ##f'## doesn't jump! ##f## is differentiable, so it may have kinks, but it does not have gaps. It goes roughly along the following lines:

differentiability of ##f## > continuity of ##f## > (Weierstraß) maximal point ##x_0## > (Fermat) ##f'(x_0)=0## > application on ##g:=f-c## where ##c## is the desired intermediate value for ##g##
 
  • Like
Likes feynman1
  • #7
fresh_42 said:
##f## is differentiable, so it may have kinks, but it does not have gaps.
why can f' not have any gaps while f' can have kinks?
 
  • #8
feynman1 said:
why can f' not have any gaps while f' can have kinks?
This is a consequence, not a cause. However, it illustrates why it works. It's a heuristic.

You take a value in the gap and construct a contradiction to Darboux. It is a bit messy with all the theorems that basically look like one another.

The reference I found is not really well written, but if you like then have a look:
https://pnp.mathematik.uni-stuttgart.de/iadm/Weidl/ana1-13-14/skript/skriptse41.html
It's in the wrong language but if you open it with Chrome you can use its translation function. Not perfect, but it should do.
 
  • Like
Likes jim mcnamara and feynman1
  • #9
I think your takeaway should be that the derivative is *basically* continuous. Do you know an example where it is not? I think it's pretty instructive to just meditate on one for a while.
 
  • #10
Office_Shredder said:
I think your takeaway should be that the derivative is *basically* continuous. Do you know an example where it is not? I think it's pretty instructive to just meditate on one for a while.
is there any counter example?
 
  • #11
fresh_42 said:
This is a consequence, not a cause. However, it illustrates why it works. It's a heuristic.

You take a value in the gap and construct a contradiction to Darboux. It is a bit messy with all the theorems that basically look like one another.
yes, is there any intuition to understand this consequence
 
  • #12
feynman1 said:
is there any counter example?

##x^2\sin(1/x)##

Basically the idea is that it is bounded by a parabola, so the derivative at ##x=0## is 0, but it wiggles around really fast at you get near 0, so you can get not close to 0 derivatives no matter how close to 0 you get.
 
  • Like
Likes fresh_42
  • #13
Office_Shredder said:
##x^2\sin(1/x)##

Basically the idea is that it is bounded by a parabola, so the derivative at ##x=0## is 0, but it wiggles around really fast at you get near 0, so you can get not close to 0 derivatives no matter how close to 0 you get.
this example is to show that its derivative jumps or not?
 
  • #14
feynman1 said:
this example is to show that its derivative jumps or not?

No, the derivative does not jump, but it is discontinuous at x=0. You should try graphing this function and its derivative and thinking for a while about what is going on.
 
  • #15
Office_Shredder said:
No, the derivative does not jump, but it is discontinuous at x=0. You should try graphing this function and its derivative and thinking for a while about what is going on.
Do you mean although the derivative is discontinuous, the derivative is always able to take any intermediate values demanded?
For a discontinuous function, the function won't in general be able to take any intermediate values demanded unless specially designed.
 
  • #16
feynman1 said:
Do you mean although the derivative is discontinuous, the derivative is always able to take any intermediate values demanded?
For a discontinuous function, the function won't in general be able to take any intermediate values demanded unless specially designed.

Have you looked at the derivative of this function near 0? Draw a graph, and evaluate the derivative at 0 yourself.
 
  • #17
Office_Shredder said:
Have you looked at the derivative of this function near 0? Draw a graph, and evaluate the derivative at 0 yourself.
yes, the derivative is discontinuous at x=0
 
  • #18
feynman1 said:
yes, the derivative is discontinuous at x=0

So this is what the darboux theorem is saying. The only way a derivative can be discontinuous is it can wiggle around and take a bunch of different values near x=a, but still be defined at x=a
 
  • #19
Office_Shredder said:
So this is what the darboux theorem is saying. The only way a derivative can be discontinuous is it can wiggle around and take a bunch of different values near x=a, but still be defined at x=a
why is this feature not in continuous functions?
 
  • #20
feynman1 said:
why is this feature not in continuous functions?
If you "differentiate" the continuous function ##x \longmapsto |x|## then you get a gap of ##+2## at zero. But if we start with a differentiable function ##f##, say at ##0##, then we have
$$
f(h)=f(0) + \varphi (h) + r(h) = f(0)+c\cdot h +r(h)
$$
with a small remainder term ##r(h)## and a linear function ##\varphi \, : \,x\longmapsto c\cdot x .## Now ##c\cdot h## is still small for small ##h## and our function value ##f(h)## is still close to ##f(0)##. I.e. there is no way to get a gap of ##+2## for small and smaller ##h.## We can have as derivative ##\varphi (x)=cx## from the left and ##\varphi' (x)=c'x## from the right, which is discontinuous if ##c\neq c'.## This means we have a discontinuity of the derivative. However, ##ch## and ##c'h## are close for small ##h##.
 
Last edited:
  • #21
feynman1 said:
why is this feature not in continuous functions?

This is the definition of continuity, that ##f(x)\approx f(a)## if ##x\approx a##.

If you have a sequence of points ##x_n## that converge to ##a##, then if ##f## is continuous we must have ##f(x_n)\to f(a)##But for the function I described above, you can find a sequence ##x_n\to 0## such that ##f'(x_n)=1/2##, yet ##f'(0)=0##.

In some sense, it would be nicer if ##f'(0)## didn't exist, because then the derivative would be continuous on its domain.
 
  • #22
fresh_42 said:
If you "differentiate" the continuous function ##x \longmapsto |x|## then you get a gap of ##+2## at zero. But if we start with a differentiable function ##f##, say at ##0##, then we have
$$
f(h)=f(0) + \varphi (h) + r(h) = f(0)+c\cdot h +r(h)
$$
with a small remainder term ##r(h)## and a linear function ##\varphi \, : \,x\longmapsto c\cdot x .## Now ##c\cdot h## is still small for small ##h## and our function value ##f(h)## is still close to ##f(0)##. I.e. there is no way to get a gap of ##+2## for small and smaller ##h.## We can have as derivative ##\varphi (x)=cx## from the left and ##\varphi' (x)=c'x## from the right, which is discontinuous if ##c\neq c'.## This means we have a discontinuity of the derivative. However, ##ch## and ##c'h## are close for small ##h##.
but when talking about discontinuity in a derivative, it is about any gap in c rather than ch? so in your example if c is different from c' the derivative is discontinuous?
 
  • #23
feynman1 said:
but when talking about discontinuity in a derivative, it is about any gap in c rather than ch? so in your example if c is different from c' the derivative is discontinuous?
Yes. My example has (excuse the sloppy notation): ##f'(0^+)=c## and ##f'(0^-)=c'##. So we have a gap of ##|c'-c|.## But the function values ##f(h)## and ##f(-h)## are close to ##f(0)## because the derivative is multiplied by ##\pm h.## This means that there is still some control on ##f(x)## by the simple fact that it is differentiable. A property we do not have for only continuous functions.
 
  • #24
Thank you, but this darboux thing says any intermediate derivative can be taken, not its function value. How to understand that?
 
  • #25
@feynman1 , you may be interested in conway's base 13 function, which is nowhere continuous but satisfies the MVP.
 
  • #26
feynman1 said:
Thank you, but this darboux thing says any intermediate derivative can be taken, not its function value. How to understand that?
You can take a look at the proof of the Darboux Theorem. One proof that you can find on Wikipedia uses two theorems: A continuous function on a closed interval achieves a maximum and a minimum and the derivative of a differentiable function at a local extremum is zero.

This second theorem importantly does not require the derivative to be continuous.
 
  • #27
feynman1 said:
Thank you, but this darboux thing says any intermediate derivative can be taken, not its function value. How to understand that?

A continuous function has the following properties:
- doesn't jump
- doesn't wiggle around too much

The derivative satisfies the doesn't jump property, but doesn't satisfy the "doesn't wiggle around too much" property.
 

Related to To understand the Darboux theorem

1. What is the Darboux theorem?

The Darboux theorem, also known as the intermediate value theorem, is a fundamental theorem in calculus that states that if a function is continuous on a closed interval, then it takes on every value between the function's values at the endpoints of the interval.

2. Why is the Darboux theorem important?

The Darboux theorem is important because it provides a way to prove the existence of solutions to equations and inequalities. It also serves as the basis for many other important theorems in calculus, such as the mean value theorem and the fundamental theorem of calculus.

3. Who discovered the Darboux theorem?

The Darboux theorem was first proved by French mathematician Jean Gaston Darboux in the late 19th century. However, the concept of the theorem has been known since ancient times, with early versions appearing in the works of mathematicians such as Archimedes and Bhaskara II.

4. What are some real-life applications of the Darboux theorem?

The Darboux theorem has many real-life applications, such as in physics, engineering, and economics. It can be used to model and analyze various phenomena, such as the motion of objects, the flow of fluids, and the behavior of financial markets.

5. Is the Darboux theorem always true?

Yes, the Darboux theorem is always true because it is a fundamental property of continuous functions. However, it is important to note that the theorem only applies to continuous functions, so it may not hold for discontinuous functions.

Similar threads

  • Calculus
Replies
12
Views
863
Replies
9
Views
467
Replies
4
Views
287
Replies
3
Views
305
  • Calculus
Replies
5
Views
1K
Replies
9
Views
1K
  • Science and Math Textbooks
Replies
3
Views
1K
  • Calculus
Replies
12
Views
2K
Back
Top