MHB Proof of Darboux's Theorem (IVT for Derivatives)

[Math Amateur]

I am reading Manfred Stoll's book: Introduction to Real Analysis.

I need help with Stoll's proof of the Intermediate Value Theorem (IVT) for Derivatives (Darboux's Theorem).

Stoll's statement of the IVT for Derivatives and its proof read as follows:
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In the above proof, Stoll argues that because

(i) there exists an $$x_1 \gt a$$ such that $$g(x_1) \lt g(a)$$

and

(ii) there exists an $$x_2 \lt $$b such that $$g(x_2) \lt g(b) $$

that as a consequence, g has an absolute minimum at some point $$c \in (a,b)$$.
If you draw some sketches, this seems a reasonably intuitive conclusion to draw ... but what is the formal, rigorous argument for this conclusion? What result(s) in analysis is Stoll drawing on and how exactly does the consequence above follow ...Hope someone can help ...Peter

 

[Fallen Angel]

Hi Peter,

The sign of the derivative tells you if the function is increasing or decreasing.

In this case $g'(a)<0$ means that in a neighbourhood of $a$, $g$ is decreasing, so you got the existence of $x_1$.

And symmetrically for $x_2$

 

[Math Amateur]

Hi Peter,

The sign of the derivative tells you if the function is increasing or decreasing.

In this case $g'(a)<0$ means that in a neighbourhood of $a$, $g$ is decreasing, so you got the existence of $x_1$.

And symmetrically for $x_2$

Hi Fallen Angel,

Thanks for the help ... but I was less concerned about the existence of $$x_1$$ and $$x_2$$ than how we could (formally and rigorously) show that this means that g has an absolute minimum at some point $$c \in (a,b)$$. (Note that I am interested in rigorously establishing the existence of an absolute minimum ... ... I think there could be any number of heuristic and intuitive arguments ... ... but i cannot frame a formal and rigorous argument to establish the conclusion.)
By the way, you write:

" ... ... $g'(a)<0$ means that in a neighbourhood of $a$, $g$ is decreasing ... ... "Is this correct?

See the remarks following Stoll's Theorem 5.2.9:
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In the above "Remarks", Exercise 18 is mentioned.

For your interest, here is the relevant Exercise 18:

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I would be most interested in your thoughts on this matter ... ...

Thanks again for your help ...

Peter

 

[Fallen Angel]

Hi Peter,

Sorry, I was missthinking that $g'$ was continuous, but you got a similar argument for the existence of $x_1$ and $x_2$ there without using continuity.

For the minimum, the points $x_1$ and $x_2$ can be chosen as near to $a$ and $b$ as needed, because you got $g(a)<g(x)$ for all $x\in (a,a+\epsilon)$ for some $\epsilon>0$ and similarly for $b$, now the interval $[x_1 ,x_2]$ contains a minimum by Weirstrass theorem.

 

[johng1]

Peter,
What you're missing is exercise 15:

If $f^\prime(a)>0$ then there is $\delta>0$ such that if $x\in(a-\delta,a)$, then $f(x)<f(a)$ and if $x\in(a,a+\delta)$, then $f(x)>f(a)$.

Proof. Apply the definition of limit. Set $\epsilon=f^\prime(a)/2$. Then there is $\delta>0$ such that if $0<|x-a|<\delta$, then $$f^\prime(a)/2<{f(x)-f(a)\over x-a}<3f^\prime(a)/2$$
In particular, for $0<|x-a|<\delta$, $$0<{f(x)-f(a)\over x-a}$$.
Now multiply the above inequality by $x-a$ to get the conclusion.

Of course, a similar statement holds if $f^\prime(a)<0$.

 

[Math Amateur]

Peter,
What you're missing is exercise 15:

If $f^\prime(a)>0$ then there is $\delta>0$ such that if $x\in(a-\delta,a)$, then $f(x)<f(a)$ and if $x\in(a,a+\delta)$, then $f(x)>f(a)$.

Proof. Apply the definition of limit. Set $\epsilon=f^\prime(a)/2$. Then there is $\delta>0$ such that if $0<|x-a|<\delta$, then $$f^\prime(a)/2<{f(x)-f(a)\over x-a}<3f^\prime(a)/2$$
In particular, for $0<|x-a|<\delta$, $$0<{f(x)-f(a)\over x-a}$$.
Now multiply the above inequality by $x-a$ to get the conclusion.

Of course, a similar statement holds if $f^\prime(a)<0$.

Thanks Johng ... Most helpful indeed ...

Peter

*** EDIT ***

Hi johng ... still reflecting ...

... can you help me with how Exercise 15 relates to providing an argument that because:

(i) there exists an $$x_1 \gt a$$ such that $$g(x_1) \lt g(a)$$

and

(ii) there exists an $$x_2 \lt $$b such that $$g(x_2) \lt g(b) $$

that as a consequence, g has an absolute minimum at some point $$c \in (a,b)$$.

I am still a bit puzzled about the rigorous justification for the above argument ... sorry if I am being slow and lacking insight ... :(

Can you help further ...

Peter

 

[johng1]

Peter,
Are you still unsure why $x_1$ and $x_2$ exist? Apply my previous post to $g^\prime(a)$ and $g^\prime(b)$. For $x_2$, choose any $x_2\in(b-\delta,b)$. Similarly for $x_1$.
Now since $g$ is differentiable on I, it is continuous on $[a,b]$. You need the theorem that says a continuous function $g$ on a closed interval $[a,b]$ has an absolute minimum; i.e. there is $x_0\in[a,b]$ such that $g(x_0)\leq g(x)$ for all $x\in[a,b]$.
Now for the case at hand could $x_0$ be $a$? No, since $g(x_0)\leq g(x_1)<g(a)$ (if it were a, you'd get $g(a)<g(a)$, absurd). Similarly, $x_0\neq b$. So $x_0\in(a,b)$.
Since $g$ is differentiable, you know $g^\prime(x_0)=0$. QED.

 

[Math Amateur]

Peter,
Are you still unsure why $x_1$ and $x_2$ exist? Apply my previous post to $g^\prime(a)$ and $g^\prime(b)$. For $x_2$, choose any $x_2\in(b-\delta,b)$. Similarly for $x_1$.
Now since $g$ is differentiable on I, it is continuous on $[a,b]$. You need the theorem that says a continuous function $g$ on a closed interval $[a,b]$ has an absolute minimum; i.e. there is $x_0\in[a,b]$ such that $g(x_0)\leq g(x)$ for all $x\in[a,b]$.
Now for the case at hand could $x_0$ be $a$? No, since $g(x_0)\leq g(x_1)<g(a)$ (if it were a, you'd get $g(a)<g(a)$, absurd). Similarly, $x_0\neq b$. So $x_0\in(a,b)$.
Since $g$ is differentiable, you know $g^\prime(x_0)=0$. QED.

Thanks for that post johng ...

That explains the issue fully and very clearly ... most helpful indeed ...

Peter