Derivatives have IVP (Romania 1996)

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Derivatives have IVP (Romania 1996)

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Source: Romanian National Olympiad 1996 – Grade 11 – Problem 1

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#1 h Apr 13, 2025, 11:29 AM

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Let $I \subset \mathbb{R}$ be a nondegenerate interval and $f:I \to \mathbb{R}$ a differentiable function. We denote $J= \left\{ \frac{f(b)-f(a)}{b-a} : a,b \in I, a<b \right\}.$ Prove that:

$a)$ $J$ is an interval;
$b)$ $J \subset f'(I),$ and the set $f'(I) \setminus J$ contains at most two elements;
$c)$ Using parts $a)$ and $b),$ deduce that $f'$ has the intermediate value property.

This post has been edited 2 times. Last edited by Filipjack, Apr 13, 2025, 11:44 AM

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#3 h Apr 17, 2025, 8:17 AM • 1 Y

Y by LawofCosine

part a:
The interval $I$ might be closed or might have one sided open interval, or may be both sided open interval ....

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#4 h Apr 17, 2025, 9:34 AM

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@above: $f'$ doesn't have to be continuous! If it were, the problem would have been pointless, because it is known that continuous functions have IVP. Find here some examples of functions with discontinuous derivative.

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#5 h Apr 17, 2025, 5:14 PM

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Filipjack wrote:

@above: $f'$ doesn't have to be continuous! If it were, the problem would have been pointless, because it is known that continuous functions have IVP. Find here some examples of functions with discontinuous derivative.

There is no need for f' is continuous to have darboux property....ya but I have to write a proof ig

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#6 h Apr 17, 2025, 7:37 PM • 2 Y

Y by Rohit-2006, Hairy_Ball_Theorem

$a)$ It is enough to prove that for any $u,v \in J$ we have $[u,v] \subseteq J$ (i.e. that $J$ is convex). Details

Let $u,v \in J.$ Then there exist $a,b,c,d$ such that $u=\frac{f(b)-f(a)}{b-a}$ and $v= \frac{f(d)-f(c)}{d-c}.$ We can assume WLOG $a<b$ (otherwise swap them) and likewise $d<c.$ Consider the function $g:[0,1] \to \mathbb{R},$ $g(t)=\frac{f((1-t)b+td)-f((1-t)a+tc)}{(1-t)(b-a)+t(d-c)}.$ This function is continuous and we have $g(0)=u$ and $g(1)=v,$ so $[u,v] \subseteq g([0,1]) \subseteq J,$ by IVT.

$b)$ For any $a<b$ in $I$ there is $c \in I$ such that $f'(c)= \frac{f(b)-f(a)}{b-a},$ by MVT. This shows that $\frac{f(b)-f(a)}{b-a} \in f'(I)$ for any $a,b \in I,$ so $J \subseteq f'(I).$

Let $p= \inf J$ and $q= \sup J.$ For any $s \in I$ we have $f'(s)= \lim_{x \to s} \frac{f(x)-f(s)}{x-s},$ but $p \le \frac{f(b)-f(a)}{b-a} \le q,$ for any $a,b \in I,$ so $p \le f'(s) \le q.$ This shows that $f'(I) \subseteq [p,q].$ Combining this with $(p,q) \subseteq J$ (established in "Details" above), we get $f'(I) \setminus J \subseteq \{p,q\}.$

$c)$ Part $b)$ shows that $f'(I)$ is an interval (to be more precise, $f'(I)$ is an interval with endpoints $p$ and $q$ ). The restriction of $f$ to any subinterval is still differentiable, so, applying the result from $b)$ to these restrictions of $f,$ we get that $f'$ transforms intervals into intervals. This trivially implies IVP (in fact it is equivalent to it).

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#7 h Apr 18, 2025, 4:49 AM

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Filipjack wrote:

$a)$ It is enough to prove that for any $u,v \in J$ we have $[u,v] \subseteq J$ (i.e. that $J$ is convex). Details

Let $u,v \in J.$ Then there exist $a,b,c,d$ such that $u=\frac{f(b)-f(a)}{b-a}$ and $v= \frac{f(d)-f(c)}{d-c}.$ We can assume WLOG $a<b$ (otherwise swap them) and likewise $d<c.$ Consider the function $g:[0,1] \to \mathbb{R},$ $g(t)=\frac{f((1-t)b+td)-f((1-t)a+tc)}{(1-t)(b-a)+t(d-c)}.$ This function is continuous and we have $g(0)=u$ and $g(1)=v,$ so $[u,v] \subseteq g([0,1]) \subseteq J,$ by IVT.

$b)$ For any $a<b$ in $I$ there is $c \in I$ such that $f'(c)= \frac{f(b)-f(a)}{b-a},$ by MVT. This shows that $\frac{f(b)-f(a)}{b-a} \in f'(I)$ for any $a,b \in I,$ so $J \subseteq f'(I).$

Let $p= \inf J$ and $q= \sup J.$ For any $s \in I$ we have $f'(s)= \lim_{x \to s} \frac{f(x)-f(s)}{x-s},$ but $p \le \frac{f(b)-f(a)}{b-a} \le q,$ for any $a,b \in I,$ so $p \le f'(s) \le q.$ This shows that $f'(I) \subseteq [p,q].$ Combining this with $(p,q) \subseteq J$ (established in "Details" above), we get $f'(I) \setminus J \subseteq \{p,q\}.$

$c)$ Part $b)$ shows that $f'(I)$ is an interval (to be more precise, $f'(I)$ is an interval with endpoints $p$ and $q$ ). The restriction of $f$ to any subinterval is still differentiable, so, applying the result from $b)$ to these restrictions of $f,$ we get that $f'$ transforms intervals into intervals. This trivially implies IVP (in fact it is equivalent to it).

I just learnt this multivariate to univariate conversion technique from a ISI senior at 2:30 a.m. but my heart went to hell when I was preparing for submitting the solution and see you already posted the solution....

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