# To understand the Darboux theorem

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feynman1
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?

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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?

feynman1
This is to compare intermediate value thm and Darboux thm. Clear?

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This is to compare intermediate value thm and Darboux thm. Clear?
Why is this property ...
[Darboux: ##f'## takes every value between ##f'(a)## and ##f'(b)##]
... 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?

feynman1
feynman1
No. Why does a derivative function possesses a property of taking any intermediate value without declaring the derivative is continuous?

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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##

feynman1
feynman1
##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?

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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:
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.

jim mcnamara and feynman1
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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.

feynman1
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?

feynman1
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

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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.

fresh_42
feynman1
##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?

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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.

feynman1
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.

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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.

feynman1
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

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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

feynman1
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?

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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##.

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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.

feynman1
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?

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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.

feynman1
Thank you, but this darboux thing says any intermediate derivative can be taken, not its function value. How to understand that?

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@feynman1 , you may be interested in conway's base 13 function, which is nowhere continuous but satisfies the MVP.

Gold Member
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.

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