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Consider the following system: $$ \begin{cases} x_1 + 3 x_3 = 4a, \\ f(x_1) + 3 f(x_3) = 8 f(a), \\ f'(x_1) = 3 f'(x_3). \end{cases} $$

I want to find all functions (or at least learn some properties that hold for all of them) $f : [0,1] \to [0,1]$ that are continuous, differentiable on $[0,1]$, monotonically decreasing on $[0,1]$ and the aforementioned system has a solution for every $a \in (0,1)$. In other words, if $a$ is fixed, there should exist $x_1, x_3 \in [0,1]$ that satisfy the system.

Or, a bit re-phrased: find $f$ such that for all $a \in (0,1)$, the aforementioned system has at least one pair $(x_1, x_3) \in [0,1]^2$ satisfying the system. And the general question is of course to find all such $f$.

UPD: Apparently, the solution is trivial only. I created another question that is hopefully more of interest: Solve differential system with a parameter

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  • $\begingroup$ I don't understand the assumptions. $x_1,x_3 \in [0,1]$ variable and the second equality is $f(x_1) + 3f(x_3) = 8f((x_1+3x_3)/4)$. Or is $x$ some constant and then the last equations leads to $f \equiv c$ for some constant $c$. So: what is variable, what is constant? $\endgroup$ Mar 4 at 19:19
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    $\begingroup$ @DieterKadelka $f$ must be such that for each $x\in(0,1)$ there exist $x_1,x_3\in[0,1]$ satisfying all three equalities. $\endgroup$ Mar 4 at 19:22
  • $\begingroup$ @DieterKadelka yes, $x$ is a parameter. Perhaps, I will rename it for more clearance $\endgroup$ Mar 4 at 20:20
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Assume such an $f$ exists. For each $a\in(0,1)$ fix some solution $x_1(a),x_3(a)$ of the system. The first equality can be restated as $$ \frac{1}{4}x_1(a) + \frac{3}{4}x_3(a) = a.$$ In other words, $a$ is a convex combination of $x_1(a),x_3(a)$. Consider a sequence $a_k \searrow 0$. As $x_1(a_k),x_3(a_k) \geq 0$ it follows that $x_1(a_k),x_3(a_k) \to 0$. Taking the limit $k\to \infty$ in the second equation, we obtain from the continuity of $f$ that $$ 4 f(0) = 8 f(0).$$ Hence $f(0) = 0$. As $f$ is monotonically decreasing and takes nonnegative values, it follows that $f$ is constant.

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