# $\int_0^x f(t)dt=\int_0^{ax}f(t)dt+ \int_0^{bx}f(t)dt$ implies $f$ constant

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 $$a,b \in (0,1)$$ $$a+b=1$$

• 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. – John Omielan Mar 13 at 21:05
• 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. – John Omielan Mar 13 at 21:08
• 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. – John Omielan Mar 13 at 21:24

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.

• Kavi.Please elaborate: We get f(ac)=f(c)?Thanks. – Peter Szilas Mar 13 at 13:27
• Already it was said the derivative equation which you and OP use is wrong. – Takahiro Waki Mar 13 at 13:41
• This is also the original solution of the problem, which I couldn't figure out how it works. Why $f(ac)=f(c)$ ? – Gaboru Mar 13 at 14:00
• Can you explain me, please? – Gaboru Mar 13 at 17:19
• @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. – Kavi Rama Murthy Mar 13 at 23:48

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.

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

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

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.