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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.