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Source: Romanian District Olympiad 2025 12.3

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#1 h Mar 8, 2025, 12:18 PM

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Let $a<b$ and $f:[a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) dx=0$ . Show that $f(a)\cdot f(b)<0$ .
Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a scrictly monotonous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $\int_{a_{n-1}}^{a_n} f(x)dx = \int_{a_n}^{a_{n+1}} f(x)dx,\text{ for all }n\geq 2.$

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#2 h Mar 10, 2025, 7:32 AM

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For $(a)$ . Otherwise, $f(a)f(b)\geq0$ , which implies $f(b)>f(a)\geq0$ or $f(a)<f(b)\leq0$ . If $f(b)>f(a)\geq0$ , then $f(x)>0,\forall x\in(a,b]$ and
$0=\int_a^bf(x)dx\geq\int_{\frac{a+b}2}^bf(x)dx\geq\int_{\frac{a+b}2}^bf\left(\frac{a+b}2\right)dx=f\left(\frac{a+b}2\right)\cdot\frac{b-a}2>0,$ which is impossible.

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#3 h Mar 12, 2025, 10:34 PM

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for $(a.)$ another easy and cute method would be, make 3 cases
$\text{Case 1:}$ $f(a) > f(b) \geq 0$
$\text{Case 2:}$ $f(a)<f(b) \leq 0$
$\text{Case 3:}$ $f(a)<0<f(b)$
$\text{for case 1 and 2 we can get contradiction and we can show case 3 works}$

for $b.$ my guess is constant sequence but i cant get anything to prove it

edit: i did it

:wow:

$\textbf{Sketch:}$ If $f$ is a strictly monotonic function, then if
$\int_{a}^{b} f(x) \,dx = 0,$ we have
$f(a) f(b) < 0.$
$\int_{a_i}^{a_j} f(x) \,dx = 0 \text{ for any } a_i < a_j.$
Taking $a_i < a_j < a_k$ , we get
$f(a_i) f(a_j) < 0, \quad f(a_i) f(a_k) < 0, \quad f(a_j) f(a_k) < 0$ which implies ( $\rightarrow \leftarrow$ ).

I initially thought that only constant sequences would work, but $\textbf{@rosemarys\_baby}$ pointed out another type of sequence: sequences that are constant up to some $n$ , differ in the next term, and remain constant from there.

A counterexample to the claim that only constant sequences work is:
counter example

This post has been edited 9 times. Last edited by Levieee, Mar 13, 2025, 5:03 PM

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#4 h Mar 13, 2025, 12:45 AM

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Idk if it's ok, otherwise, pls tell me the mistake.
WLOG, consider an increasing function. The other case is analogous.
a)Using Rolle's Theorem in $F(x)$ , there exist a $k$ such $f(k)=0$ . If $a<k<b$ we are done.
If $k \le a<b$ , $0 \le f(a)<f(b) \Rightarrow 0 \le f(x) \Rightarrow 0<\int^a_b f(x) dx$ because is strictly monotonous and positive function. Contradiction. Analogous, $a < b \le 0$ .
b)Suppose that $(a_n)_{n \ge 1}$ works. Let $b_n=\int^{a_{n}}_{a_{n-1}}f(x)dx$ . As $b_2=b_3=...$ , $(b_n)$ converges. Let $L_b=\lim_{n \rightarrow \infty} b_n$ and $L_a=\lim_{n \rightarrow \infty }a_n$ .
Notice that $L_b=\lim_{n \rightarrow \infty} b_n =\lim_{n \rightarrow \infty} \int^{a_{n}}_{a_{n-1}}f(x)dx = \int^{L_a}_{L_a}f(x)dx =0$ .
As $b_2=b_3=...$ , we have that $b_n=L_b=0$ for $n>1$ .
Supose that there exist $a_i, a_j, a_k$ such $a_i < a_j <a_k$ Using (a): $f(a_i)f(a_j)<0$ and $f(a_j)f(a_k)<0$ . But, as $f$ is strictly monotonous function, $f(a_i)<f(a_j)<f(a_k)$ . Contradiction. So $(a_n)$ can be at most two possible values. Let's call them $c, d \in \mathbb{R}$ . As $a_n$ converges, WLOG, there exist a $N$ such that, for all $n \ge N$ , $a_n=c$ .
If $a_n$ is a sequence such that $a_n \in \{c, d\}$ and there exist a $N$ such that, for all $n \ge N$ , $a_n=c$ , we have got two possibilities:
-If $a_i=a_{i+1}$ , $\int_{a_{i}}^{a_{i+1}} f(x)dx=0$
-If $a_i=c, a_{i+1}=d$ , consider an increasing linear function $f(x)$ that pass through the midpoint of $CD$ (let's call it $m$ ) and $(c, f(c)); (d, f(d))$ (let's call them $C', D'$ , respectively) are the intersection of the linear function and the perpendicular to the x-axis that pass through $c$ and $d$ , respectively. By $ALA$ , easy to see that $\triangle MCC' \cong \triangle MDD'$ , so $\int_{a_{i}}^{a_{i+1}} f(x)dx=0$ .
With that construction, the recurrence works.
Answer: All sequences that $a_n \in \{c, d\}$ and there exist a $N$ such that, for all $n \ge N$ , $a_n=c$ .

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#5 h Mar 13, 2025, 1:07 AM

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@above: in part a) you cannot use Rolle's Theorem because the function $\int\limits_c^x f(t) \mathrm{d}t$ is not necessarily differentiable. It is however continuous, so your solution for part b) is fine (this was important for the calculation of $\lim_{n \to \infty} b_n$ ).

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#6 h Mar 13, 2025, 1:23 AM • 1 Y

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Filipjack wrote:

@above: in part a) you cannot use Rolle's Theorem because the function $\int\limits_c^x f(t) \mathrm{d}t$ is not necessarily differentiable. It is however continuous, so your solution for part b) is fine (this was important for the calculation of $\lim_{n \to \infty} b_n$ ).

doesn't FTOC state that if a function is continuous at $c_{1}$ then $F(x)$ must be differentiable at $c_{1}$ and $F'(c_{1})=f(c_{1})$ $c_{1} \in [a,b]$

@below my bad i didnt read that continuity wasn't given

This post has been edited 2 times. Last edited by Levieee, Mar 13, 2025, 12:59 PM

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#7 h Mar 13, 2025, 11:26 AM • 1 Y

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Levieee wrote:

doesn't FTOC state that if a function is continuous at $c_{1}$ then $F(x)$ must be differentiable at $c_{1}$ and $F'(c_{1})=f(c_{1})$ $c_{1} \in [a,b]$

The thing is: we are not given that $f$ is continuous, so you cannot apply that.

Take for example $f(x)= \begin{cases} x-1, & x \le 0 \\ x+1, & x>0 \end{cases}$ and $F(x)=\int\limits_c^x f(t) \mathrm{d}t,$ for some $c.$ For this we have $F(x)=x^2+|x|-c^2-|c|,$ which is not differentiable in $0.$ Notice also that there is no point $s$ such that $f(s)=0.$

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#9 h Mar 13, 2025, 5:19 PM

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KAME06 wrote:

Idk if it's ok, otherwise, pls tell me the mistake.

Your answer for part (b) matches with mine. So maybe it's correct. Will add mine a bit later.

@OP there's a little typo in your problem statement. FTFY.
code

[list=a]
[*] Let $a<b$ and $f\colon [a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) \mathrm d x=0$. Show that $f(a)\cdot f(b)<0$.
[*] Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a strictly monotonous function $f\colon\mathbb{R}\rightarrow\mathbb{R}$ such that \[\int_{a_{n-1}}^{a_n} f(x)\mathrm dx = \int_{a_n}^{a_{n+1}} f(x)\mathrm dx,\text{ for all }n\geq 2.\]
[/list]

Let $a<b$ and $f\colon [a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) \mathrm d x=0$ . Show that $f(a)\cdot f(b)<0$ .
Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a strictly monotonous function $f\colon\mathbb{R}\rightarrow\mathbb{R}$ such that $\int_{a_{n-1}}^{a_n} f(x)\mathrm dx = \int_{a_n}^{a_{n+1}} f(x)\mathrm dx,\text{ for all }n\geq 2.$

This post has been edited 3 times. Last edited by kamatadu, Mar 13, 2025, 8:47 PM

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#10 h Mar 13, 2025, 7:42 PM

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Solution.

Part (a).

Note that the conditions remain same if we consider $-f$ instead of $f$ . So WLOG assume $f$ is strictly increasing. Therefore $f(a)<f(b)$ .

Case 1: $f(a)=0$ .

Then,
$\begin{align*} \int_a^b f(x)\mathrm{d}x &= \int_a^{\frac{a+b}{2}}f(x)\mathrm{d}x +\int_{\frac{a+b}{2}}^b f(x)\mathrm{d}x \\ &\ge \int_a^{\frac{a+b}{2}}f(a)\mathrm{d}x +\int_{\frac{a+b}{2}}^b f\left(\frac{a+b}{2}\right)\mathrm{d}x \\ &= \frac{b-a}{2}\cdot f(a)+\frac{b-a}{2}\cdot f\left(\frac{a+b}{2}\right)>0 . \end{align*}$
But we know that $\int_a^b f(x)\mathrm{d}x=0$ which gives us our contradiction.
Case 2: $f(b)=0$ .

This follows similarly like the previous case.
Case 3: $f(a)$ , $f(b)\neq 0$ .
$0=\int_a^b f(x)\mathrm{d}x \ge \int_a^b \mathrm{d}x =f(a)\cdot \left(\frac{b-a}{2}\right) .$ Similarly, we also get that,
$f(b)\cdot \left(\frac{b-a}{2}\right) \ge 0 .$ Multiplying these two, we get,
$f(a)f(b)\left(\frac{b-a}{2}\right)^2\le 0 \implies f(a)\cdot f(b)<0 .\hfill\square$

Part (b).

Let,
$\lambda = \int_{a_1}^{a_2} f(x)\mathrm{d}x =\int_{a_2}^{a_3} f(x) \mathrm{d}x = \cdots .$ Therefore, $n\lambda = \int_{a_1}^{a_2} f(x)\mathrm{d}x +\cdots+ \int_{a_n}^{a_{n+1}} f(x)\mathrm{d}x = \int_{a_1}^{a_{n+1}} f(x)\mathrm{d}x$ .

Define $b_n = \int_{a_1}^{a_{n+1}}f(x) \mathrm{d}x$ for all $n\ge 1$ . Note, $b_{n+1}-b_n=\int_{a_{n+1}}^{a_{n+2}}f(x)\mathrm{d}x =0\implies b_{n+1}=b_n$ .

This implies that $b_n$ converges which further implies that $b_n$ is bounded. So there exists $M>0$ such that $\lvert b_n \rvert \le M$ .

So,
$\lvert \lambda \rvert \le \left\lvert \frac{b_{n+1}}{n} \right\rvert \le \left\lvert \frac{M}{n} \right\rvert \implies \lvert \lambda \rvert \le \lim_{n \to \infty} \left\lvert \frac{M}{n} \right\rvert=0 \implies \lambda = 0 .$
Claim: $\left\{ a_n \right\}_{n\ge 1}$ has at most two different terms.

Proof. FTSOC assume that $\left\{ a_n \right\}_{n\ge 1}$ has at least $3$ different terms.
Let $a_i$ , $a_j$ , $a_k$ be three such that in order, i.e., $a_i<a_j<a_k$ . Firstly note that for $m<n$ , $0 = \int_{a_m}^{a_{m+1}} f(x)\mathrm{d}x=\cdots =\int_{a_{n-1}}^{a_n} f(x)\mathrm{d}x .$ Summing them all, we get that $\int_{a_m}^{a_n}f(x) \mathrm{d}x = \int_{a_n}^{a_m}f(x)\mathrm{d}x =0$ for all $m$ , $n\in \mathbb{N}$ .
So using this, we can state that, $\int_{a_i}^{a_j} f(x)\mathrm{d}x =\int_{a_j}^{a_k} f(x)\mathrm{d}x = \int_{a_i}^{a_k} f(x)\mathrm{d}x =0 .$ Using Part (a) and our previous finding, we get that, $f(a_i)\cdot f(a_j)<0, f(a_j)\cdot f(a_k)<0, f(a_i)\cdot f(a_{k})<0 .$ Multiplying these three, we get a contradiction as, $0 > (f(a_i)\cdot f(a_j)\cdot f(a_k))^2 \ge 0 .\hfill\square$
So we can further divide this claim into two cases.

$\left\{ a_n \right\}_{n\ge 1}$ is a constant sequence: Consider $f(x)=x$ . A bit of checking shows that this works. $\hfill\square$
$\left\{ a_n \right\}_{n\ge 1}$ has exactly two different terms in it:
Consider $f(x)=\frac{2(x-b)}{b-a}+1$ . Then note that, $\int_a^b f(x)\mathrm{d}x =\int_b^a f(x)\mathrm{d}x =0$ .
If $(a_{n-1},a_n)=(a,b)$ or $(b,a)$ or $(a,a)$ or $(b,b)$ , then we have,
$\int_{a_{n-1}}^{a_n} f(x)\mathrm{d}x =0 .$ So such an $f$ works. $\hfill\square$

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#11 h Yesterday at 8:40 AM

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Part (a)
Let $M=\max{(f(a), f(b))}, \ m=\min{((f(a), f(b))}.$ Since $f$ is strictly monotone, we have $m<f(x)<M$ for all $x\in(a, b)$ .
Now suppose $f(a)f(b)\ge 0$ , i.e., $mM\ge 0$ . Then either $m, M$ are both non-negative ( $\ge 0$ ), or both non-positive ( $\le 0$ ).
Since, $m<f(x)<M$ for all $x\in(a, b)$ , we have $\int_a^b m \ dx < \int_a^b f(x) \ dx < \int_a^b M \implies (b-a)m<0 < (b-a)M\implies m< 0 < M.$ which is a contradiction as both $m, M$ are either $\ge 0$ or $\le 0$ .

Part (b)
Let $\lim_{n\rightarrow \infty} a_n = l$
Let $k = \int_{a_1}^{a_2} f(x) \ dx = \int_{a_2}^{a_3} f(x) \ dx = \cdots.$ So, $(n-1)k=\int_{a_1}^{a_2} f(x) \ dx + \int_{a_2}^{a_3} f(x) \ dx + \cdots + \int_{a_{n-1}}^{a_n} f(x) \ dx=\int_{a_1}^{a_n} f(x) \ dx.$ $\implies \lim_{n\rightarrow \infty} \int_{a_1}^{a_n} f(x) \ dx = \lim_{n\rightarrow \infty} (n-1)k$ $\implies \int_{a_1}^{l} f(x) \ dx= \lim_{n\rightarrow \infty} (n-1)k.$ If $k\neq 0$ , then $\int_{a_1}^l f(x) \ dx=\pm \infty$ . But we know $\int_{a_1}^l f(x) \ dx$ is bounded between $(l-a_1)f(a_1)$ and $(l-a_1)f(l)$ , both of which are finite (as $f$ is strictly monotone). So, $k$ must be $0$ .
So, we must have $\int_{a_n}^{a_{n+1}} f(x) \ dx = 0$ for all $n\ge 1$ .
So, we have $\int_{a_m}^{a_n} f(x) \ dx = 0$ for any $m,n\ge 1$
Now for any two reals $a<b$ , we can find a monotone function $f$ such that $\int_a^b f(x) \ dx = 0$ [just take $f(x) = x-\frac{a+b}{2}$ ].

Claim: The sequence $(a_n)_{n\ge 1}$ cannot have more than two distinct terms.
Proof: Suppose the sequence has atleast 3 distinct terms. Let $a<b<c$ be three terms of the sequence. Since, $\int_a^b f(x) \ dx = \int_b^c f(x) = \int_a^c f(x) = 0$ , we have $f(a)f(b)<0$ and $f(b)f(c)<0$ and $f(c)f(a)<0$ (from part (a)). Multiplying the three will give $(f(a)f(b)f(c))^2<0$ , which is a contradiction.

So, $(a_n)_{n\ge 1}$ must have atmost 2 distinct terms.
So, we have the following cases:

Case 1: $(a_n)_{n\ge 1}$ is a constant sequence. Clearly we have $\int_{a_i}^{a_{i+1}}f(x) \ dx = 0$ for all $i\ge 1$ and for any strictly monotone function $f$ .
Case 2: $(a_n)_{n\ge 1}$ has exactly two distinct terms $a<b$ .
We have earlier shown the existence of a strictly monotone function $f$ such that $\int_a^b f(x) \ dx = 0$ . Also, $\int_a^a f(x) \ dx = \int_b^b f(x) \ dx = 0$ . Let $S_a = \{i \ | \ a_i=a\}$ , $S_b = \{i \ | \ a_i=b\}$ .
If both $S_a$ and $S_b$ are infinite, then consider the subsequences $(a_i)_{i\in S_a}$ and $(a_i)_{i \in S_b}$ , they converge to $a$ and $b$ respectively. So, $(a_n)_{n\ge 1}$ does not converge, a contradiction.
So, one of $S_a$ and $S_b$ must be finite. In that case the sequence $(a_n)_{n\ge 1}$ will be convergent.

So, any eventually constant sequence consisting of atmost 2 distinct terms will work.

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This post has been edited 1 time. Last edited by Fibonacci_math, Yesterday at 8:42 AM

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