1. ## Darboux Theorem

This is my lecturer's Darboux Theorem, however I have a problem with one of the steps there, I signed it wih ***. He said xmin is not equal to a and b. Buf if it is equal then what? I mean intuitively when I think about it if it is equal then I cant get c such that f ' (c)=0 but still can I just say xmin is not equal to a and b? Isn't that specific case? Thank you.

2. ## Re: Darboux Theorem

Originally Posted by davidciprut
He said xmin is not equal to a and b. Buf if it is equal then what?
He explains in the folloing two lines why $x_{text{min}}$ cannot equal $a$ or $b$, though his explanation leaves a lot to be desired. A function cannot be decreasing in one point, only on a set. However, it is easy to show from the definition of derivative that if $f'(a)<0$, then there exists a $t>a$ such that $f(t)<f(a)$.

3. ## Re: Darboux Theorem

So he had to say that if f'(a)<0 then there exists h>0 such that a-h<x<a+h where the function is decreasing. so there exsist t such that a<t<a+h with f(t)<f(a). And since it's an interval you can find a point that is smaller than a and therefore it can't be xmin I get it ! Thanks

4. ## Re: Darboux Theorem

Originally Posted by davidciprut
So he had to say that if f'(a)<0 then there exists h>0 such that a-h<x<a+h where the function is decreasing.
If $f'(a)<0$, it does not follow that there exists a neighborhood of $a$ where $f$ is decreasing. That would be true if $f'(x)$ were continuous at $a$: then there would be a neighborhood where $f'(x)<0$ and we could use the mean value theorem to show that in $f$ is decreasing in that neighborhood. But $f'(x)$ does not have to be continuous.

The following example is taken from the book "Counterexamples in Analysis" (chapter 3, example 5). Let
$$f(x)= \begin{cases} x+2x^2\sin\frac{1}{x}&\text{if }x\ne0\\ 0&\text{if }x=0 \end{cases}$$
Then
$$f'(x)= \begin{cases} 1+4x\sin\frac{1}{x}-2\cos\frac{1}{x}&\text{if }x\ne0\\ 1&\text{if }x=0 \end{cases}$$
Thus $f'(1)=1>0$, but the term $2\cos(1/x)$ causes oscillations around $x=0$. The result is that $f'(x)$ takes both positive and negative values, and so $f(x)$ increases and decreases, in each neighborhood of 0. As expected, $f'(x)$ is not continuous at 0.

What I was saying is that if $f'(a)<0$, then using the definition of derivative it is easy to show that there exists (at least) one point $t>a$ such that $f(t)<f(a)$.

5. ## Re: Darboux Theorem

If h(x) is continuous on [a,b], it takes on all values between h(a) and h(b). (standard theorem)

let h(x) =f'(x)

6. ## Re: Darboux Theorem

Originally Posted by emakarov
If $f'(a)<0$, it does not follow that there exists a neighborhood of $a$ where $f$ is decreasing.
Yes, you are right, however according to the derivative definition, L=f'(a)<0 exists , which means lim(f(x)-f(a))/((x-a)) when x goes to a exists, and according to the definition of limit
For every epsilon there exists delta such that, |x-a|<delta then |f(x)-f(a))/((x-a))-L|<epsilion, and this means that there is a neighbourhood, otherwise we couldn't talk about the limit. Since f'(a) exists so does the limit and therefore according to the definition of limits there has to be a neighbourhood. Is my logic wrong?

7. ## Re: Darboux Theorem

If $f'(a)<0$, then there does indeed exist a neighborhood of $a$ where $f(x)>f(a)$ for $x<a$ and $f(a)>f(x)$ for $x>a$. However, there may not be a neighborhood of $a$ where $f(x)$ is decreasing. Do you see the difference between the two claims?

8. ## Re: Darboux Theorem

Hmmm, I can't really tell the difference. I am going to think about it for a while and if I can't tell the difference I am going to post a message again. Thank you for giving feedbacks, I really appreciate it.

9. ## Re: Darboux Theorem

My previous post was wrong. Continuity of f’(x) is sufficient but not necessary to prove Darboux’s theorem by intermediate value theorem.
So I took OP proof seriously. Having done so, the OP attachment seems quite straight-forward, and I don’t see all the fuss about ***:

The OP begins:
Let f’(a)<0<f’(b).
f(x) continuous on [a,b], therefore f(x) achieves a min on [a,b], and
a<xmin<b by definition of right and left hand derivative.*

*f(x)<f(a) for points near a to the right of a (f’(a)<0) and f(x)<f(b) for points near b to the left of b (f’(b)>0).

Continuity of derivative is irrelevant to this proof.

10. ## Re: Darboux Theorem

Originally Posted by Hartlw
*f(x)<f(a) for points near a to the right of a (f’(a)<0)
The question was how to show that $f'(a)<0$ implies that there exists a $\delta$ such that $f(x)<f(a)$ for $a<x<a+\delta$. The OP said that $f$ decreases at $a$, which does not make sense. It also does not follow that there exists a $\delta$ such that $f$ is decreasing on $[a,a+\delta]$.

11. ## Re: Darboux Theorem

Originally Posted by emakarov
The question was how to show that $f'(a)<0$ implies that there exists a $\delta$ such that $f(x)<f(a)$ for $a<x<a+\delta$. The OP said that $f$ decreases at $a$, which does not make sense. It also does not follow that there exists a $\delta$ such that $f$ is decreasing on $[a,a+\delta]$.
f’(a)=lim (f(a+h)-f(a))/h, h→0, h+
|(f(a+h)-f(a))/h – f’(a)|< ε if h<δ
-ε < (f(a+h)-f(a))/h – f’(a) < ε
f(a+h) –f(a) < h[f’(a)+ε]

f’(a)= -|f’(a)| and let ε =|f’(a)|/2
then
f(a+h) < f(a) and f(a) is not a minimum.

A similar argument applies at b.