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Let $a,b \in (0,1)$ be such that $a+b=1$ and $f:[0,1] \to \mathbb R$ be a continuous function such that $ \int_0^x f(t)dt=\int_0^{ax}f(t)dt+ \int_0^{bx}f(t)dt$. We have to prove that $f$ is constant.

Using the derivative, we get: $f(x)=af(ax)+bf(bx)$

I did the case $a=b=1/2$, but I don't know how to make it with $a,b$ arbitrary and $a,b \in (0,1)$ $a+b=1$

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  • $\begingroup$ Note the OP had already asked this question about $2$ hours earlier at Proving that a function is constant from functional equation, with this being the final corrected version of it. $\endgroup$ – John Omielan Mar 13 at 21:05
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    $\begingroup$ Please don't do this repetition of questions. I spent quite a bit of time writing an answer for you to your earlier question. If I had known then about this one existing, with $2$ solutions already, I wouldn't have bothered. In general, don't repeat the same question. If you do, or even just a closely related question, please at least provide links to each other so everybody knows about the connection. Thanks. $\endgroup$ – John Omielan Mar 13 at 21:08
  • $\begingroup$ I've chosen to move my answer to here because it might have some value, plus the other question might be closed & deleted as a duplicate. $\endgroup$ – John Omielan Mar 13 at 21:24
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Let $f$ attain its minimum at $c$. Then $f(c) =af(ac)+bf(bc) \geq af(c)+bf(c)=f(c)$. Equality must hold throughout and we get $f(ac)=f(c)$. Iterating and taking limit we get $f(c)=f(0)$. Similarly the maximum value of $f$ is also $f(0)$ . Hence $f$ is a constant.

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  • $\begingroup$ Kavi.Please elaborate: We get f(ac)=f(c)?Thanks. $\endgroup$ – Peter Szilas Mar 13 at 13:27
  • $\begingroup$ Already it was said the derivative equation which you and OP use is wrong. $\endgroup$ – Takahiro Waki Mar 13 at 13:41
  • $\begingroup$ This is also the original solution of the problem, which I couldn't figure out how it works. Why $f(ac)=f(c)$ ? $\endgroup$ – Gaboru Mar 13 at 14:00
  • $\begingroup$ Can you explain me, please? $\endgroup$ – Gaboru Mar 13 at 17:19
  • $\begingroup$ @Gaboru Sorry, I had logged out when you asked for details. You seem to have figured out the details, but let me know if you need some more clarifications. $\endgroup$ – Kavi Rama Murthy Mar 13 at 23:48
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Since $f$ is continuous, for each $\epsilon>0$, there exists $\delta>0$ such that $$0\le t\le \delta \implies |f(t)-f(0)|\le \epsilon.$$ Recursively, we have that $$\begin{align*} f(x)&=af(ax)+bf(bx) \\&=a^2f(a^2x)+2abf(abx)+b^2f(b^2x) \\&=\cdots \\&=\sum_{i=0}^n \binom{n}{i}a^ib^{n-i} f(a^ib^{n-i}x). \end{align*}$$ Note that since $\max\{a,b\}<1$, it holds $a^i b^{n-i}x\le \max\{a,b\}^n x\le \delta$ for all $0\le i\le n$ for sufficiently large $n$, which implies $$\begin{align*} |f(x)-f(0)|&\le \sum_{i=0}^n \binom{n}{i}a^ib^{n-i} |f(a^ib^{n-i}x)-f(0)|\\&\le \sum_{i=0}^n\binom{n}{i}a^ib^{n-i}\epsilon \\&=(a+b)^n\epsilon=\epsilon, \end{align*}$$ by the binomial theorem. Since $\epsilon>0$ was arbitrary, we have $f(x)=f(0)$ for all $x$ as desired.

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  • $\begingroup$ So it seems like this is also true for $a,b>0$ such that $a+b \leq 1$. $\endgroup$ – Joshhh Mar 13 at 12:58
  • $\begingroup$ Yes, I think so! But we should also note that if $a+b<1$, then $f=0$ is the only solution :) $\endgroup$ – Song Mar 13 at 13:03
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You've already determined that

$$f(x)=af(ax)+bf(bx) \tag{1}\label{eq1}$$

Assume the function $f$ is not constant. Since it's a continuous function on a closed set, the Extreme value theorem states that

$f$ must attain a maximum and a minimum, each at least once.

Choose the largest value of $x \le 1$ that is an extrema point, i.e., maximum or minimum, and call it $x_1$. Note since $f(0)$ can't be both the minimum & maximum, that $x_1 \gt 0$. Assume initially it's a maximum. By continuity and that there is a minimum point $\lt x_1$, we can choose a point $0 \lt x_2 \lt x_1$ where $f(x_2) \lt f(x_1)$. Next, note that

$$f(x_1) = af(x_1) + (1 - a)f(x_1) \tag{2}\label{eq2}$$

Let $a = \frac{x_2}{x_1}$, so $ax_1 = x_2$. Also, let $bx_1 = x_3$. Using this along with $x = x_1$ and $b = 1 - a$ in \eqref{eq1} gives

$$f(x_1) = af(x_2) + (1 - a)f(x_3) \tag{3}\label{eq3}$$

Next, \eqref{eq2} - \eqref{eq3} gives

$$0 = a(f(x_1) - f(x_2)) + (1 - a)(f(x_1) - f(x_3)) \tag{4}\label{eq4}$$

Since $a \gt 0$, $f(x_1) - f(x_2) \gt 0$ and $1 - a \gt 0$, this means that $f(x_1) - f(x_3) \lt 0$, i.e., $f(x_3) \gt f(x_1)$. However, $f(x_1)$ was the maximum, so this is not possible. Thus, in this case, the original assumption of $f$ not being constant must be false. You can repeat basically the same arguments for the case where $f(x_1)$ is a minimum instead to show that, overall, $f$ must be a constant function.

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