Problem from Zorich's book volume 1

Question

Show that if a function $f$ is defined and differentiable on an open interval $I$ and $[a,b]\subset I$, then

a) the function $f'(x)$ (even if it is not continuous!) assumes on $[a,b]$ all the values between $f'(a)$ and $f'(b)$;

b) if $f''(x)$ also exists in $(a,b)$, then there is a point $\xi\in (a,b)$ such that $f'(b)-f'(a)=f''(\xi)(b-a).$

My approach: a) I know that this is the well-known Darboux's theorem and I was able to prove it.

b) But I have issues with this part. It looks very similar to mean value theorem but MVT is not applicable here because $f'(x)$ is continuous on the open interval $(a,b)$.

I guess since it comes after Darboux's theorem then probably it could derived via part a) but I cannot see this.

Would be thankful for the solution.

If $f''(x)$ exists, then $f'$ is continuous and the regular mean value theorem applies. Are you interpreting (b) to mean that $f''$ doesn't always exist somewhere? — davidlowryduda, Apr 12 '21 at 01:13
@davidlowryduda, I do know that if $f''(x)$ exists then $f'(x)$. This is a well known fact actually. But in our case $f''(x)$ exists in $(a,b)$ which means that $f'(x)$ is continuous on $(a,b)$. In order to apply MVT to $f'(x)$ this function should be continuous on the closed interval $[a,b]$. — RFZ, Apr 12 '21 at 01:26
I agree with davidlowryduda for the suggestion for the point b); the point a) is a very famous property of the derivative, due to Darboux. I remember to have encountered the proof on the Principles of Mathematical Analysis, from Rudin. — Amelian, Apr 12 '21 at 01:47
@Amelia, I did not get his idea to be honest. But I already explained it why MVT might not work in this case. — RFZ, Apr 12 '21 at 01:49
I misread your question and thought that f was differentiable in the whole closed interval I. Now like you I think we can't apply directly the mean value theorem, sorry. Can I ask you which page from Zorich you found this problem? I would like to take a loot at it directly from the book. — Amelian, Apr 12 '21 at 01:52
@Amelia, MVT tells us that if $g(x)$ is continuous on $[a,b]$ and $g(x)$ is differentiable on $(a, b)$ then $\exists \xi \in (a, b)$ such that $g'(\xi)(b-a)=g'(b)-g'(a)$, right? But in our case we are considering $f'(x)$ instead of $g(x)$. In order to apply MVT $f'(x)$ needs to be continuous on $[a,b]$, right? But in our case it is just continuous on $(a,b)$. — RFZ, Apr 12 '21 at 01:52
Okay, last time I change my mind: since f is differentiable on I and $[a,b] \subseteq I$, then f is differentiabl on $[a,b]$, thus continuous there, and the MVT indeed apply. — Amelian, Apr 12 '21 at 01:54
All I can establish is that there is a sequence $\xi_n$ such that $f''(\xi_n) \to (f' (b) - f'(a)) / (b-a) $. Don't how this will (or will not) lead to the desired conclusion. — Paramanand Singh, Apr 12 '21 at 12:23
@ParamanandSingh: A possible continuation to your sequence: define $g(x)=f'(x)-\frac{f'(b)-f'(a)}{b-a} (x-a)$ immediately giving $g(a)=g(b)$ and so $g'(x)=f"(x)-\frac{f'(b)-f'(a)}{b-a}$ immediately resulting in $g'(\xi_n)\to 0$ which proves the result by Darboux's theorem. Although I may be wrong. — Koro, Apr 12 '21 at 13:37
@Koro: what you have done is essentially rotating graph the $f'(x) $ to get $g(x) $. This does not help. We don't get a value $\xi$ such that $g'(\xi) =0$. — Paramanand Singh, Apr 12 '21 at 13:45
@Koro: I am trying to find a counter-example, but not found so far. Maybe the problem needs a different approach. — Paramanand Singh, Apr 12 '21 at 13:47

Koro · Accepted Answer · 2021-04-12T15:21:05.063

6

For part $(b)$: Suppose on the contrary that no such $\xi$ exists then for all $t\in (a,b)$, we have
$f''(t)\ne \frac{f'(b)-f'(a)}{b-a}$

If for some $t_1,t_2\in (a,b)$ such that $t_1\ne t_2$ we have $f''(t_1)\gt \frac{f'(b)-f'(a)}{b-a}\gt f''(t_2)$ then by Darboux theorem there exists some $x_0\in (t_1,t_2)$ such that $f''(x_0)=\frac{f'(b)-f'(a)}{b-a}$ (call it $p$ for brevity), which violates our assumption. So WLOG, let $f''(x)\gt p$ for $x\in (a,b).$

Define: $g(x)=f'(x)-p(x-a)\implies g'(x)\gt 0$ for all $x\in (a,b)$
$g(a)=g(b)=f'(a)$ and noting that $g$ is strictly increasing on $(a,b)$, now $\exists c \in (a,b): g(c)\ne g(a)$. (If not then $g(x)=g(a)$ for all $x\in (a,b)$)

WLOG, let $g(c)\gt g(a)=g(b),$ (the other case is similar) by Darboux theorem $\exists c_1\in (c,b)$ such that $g(c)\gt g(c_1)\gt g(b)$ which violates that $g$ is strictly increasing on $(a,b)$.

Therefore, it is not true that $f''(x)\gt p$ for all $x\in (a,b)$.

Similarly, it is not true that $f''(x)\lt p$ for all $x\in (a,b)$. Hence by contradiction $\exists \theta \in (a,b)$ such that $f''(\theta)=p$

edited Apr 12 '21 at 15:21

answered Apr 12 '21 at 06:37

Koro

11,402

How does Darboux give us a desired $c_1$? – Paramanand Singh Apr 12 '21 at 15:24
@ParamanandSingh: Note that $g$ is a derivative and every derivative satisfies IVT. Infact you will agree that $\frac d{dx} (f(x)-p(\frac {x^2}2-ax))=g(x)$ – Koro Apr 12 '21 at 15:26
Ok got it. Another way: since $g$ is increasing it must have one sided limits at each point (including $a, b$) and that means $f'$ has limits at $a, b$ and these must be $f'(a), f'(b) $ and that makes it continuous. – Paramanand Singh Apr 12 '21 at 15:36
@ParamanandSingh: In fact, I thought about that (existence of one-sided limits) today morning, posted the answer considering existence of one-sided limits but realized it was wrong so deleted it afterwards. Consider $h(x)=\ln x$ strictly increasing on $x\gt 0$ but $\lim h(x)$ does not exist as $x\to 0$. Please correct me if I'm wrong. – Koro Apr 12 '21 at 15:40
Nice approach! +1 given there. – Paramanand Singh Apr 12 '21 at 15:44
My fault. I should have given more details. If $g$ is increasing on $(a, b) $ then limit at $a$ can also be $-\infty $ and limit at $b$ can also be $\infty$. At intermediate points both left right limits exist finitely. – Paramanand Singh Apr 12 '21 at 15:47
@ParamanandSingh: Thanks a lot I appreciate that. About the limit at end points, I think that your earlier statement is actually true on extended reals under certain conditions :) But whether a proof can be constructed using that, is an another question. – Koro Apr 12 '21 at 16:01
I don't want to write a full blown answer because the idea came by looking at your function $g$. Do try to go through my comments again and you will find it is already a complete and rigorous proof. Some details may have to be given though. – Paramanand Singh Apr 12 '21 at 16:47
Let me ask you a small question just to be sure that I am understanding it correctly: So we consider function $g$ on the $[c,b]$. Since $g(b)<g(c)$ then by Darboux's theorem we can find $c_1\in (c,b)$ such that $g(b)<g(c_1)<g(c)$ which is contradiction, right? Because $g(x)$ is increasing and since $c_1<b$ it should be $g(c_1)<g(b)$. Right? – RFZ Apr 12 '21 at 18:13
@ZFR: No. $g(c_1)\lt g(c)$ is a contradiction. Why? Because $g$ is strictly increasing on open interval $(a,b)$. – Koro Apr 12 '21 at 18:27
@Koro, Oh I see. That is correct! Thank you so much! I am about to accept it as the best answer! +1 – RFZ Apr 12 '21 at 18:42
@ZFR: You're welcome. :) – Koro Apr 12 '21 at 18:47
@Koro, Hi! Can you take a look at this topic please?https://math.stackexchange.com/questions/4117959/convex-functions-and-its-differential-properties-zorichs-book – RFZ Apr 29 '21 at 21:51
@ZFR: Hi! I’m afraid I don’t currently know much about convexity/ concavity of a function but I’ll take a look at the question. – Koro Apr 30 '21 at 06:04

Problem from Zorich's book volume 1

1 Answers1

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