How to prove that the functional equation $f(x)+f(y)=f(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)})+f(\frac{x+y}2)$ is verified only by some basic functions?

It's an question about functional analysis :

Let $x,y>0$ and $f$ a continuous and convex/concave function on $(0,\infty)$ with $f'(x)\neq 0$ on the previous interval and: $$f(x)+f(y)=f\left(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}\right)+f\left(\frac{x+y}{2}\right)$$

My claim is :

The only function verifying this equation are $$\pm (cx+b)$$ $$\pm (\frac{c}{x}+b)$$ $$\pm \ln(x)$$

It's not hard to check that the functions above verify the functional equation .

Edit :

It seems that the only function wich verify the functional above are reciprocally convex/concave .

My question : How to prove it ? Is the sentence of my edit true ?

Thanks a lot for your time .

Answer: Besides the families $f(x)=ax+b$, $f(x)=\frac a{x+b}+c$ and $f(x)=a\ln(bx+c)$ of solutions of the functional equation $$f(x)+f(y)=f\left(\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}\right)+f(\frac{x+y}2)$$ mentioned in the question, there also are the families $$f(x)=a\ln\left(\frac{x+b}{x+c}\right)+d\mbox{ with positive }b,c$$ and $$f(x)=a\arctan(bx+c)+d\mbox{ if }b\neq0.$$ This completes the list of all its solutions. It turns out, that members of the first new family are also reciprocally convex/concave, membres of the second one only if $bc\geq0$. If $bc<0$, then the members of the second new family are not even convex/concave. Observe that the only solutions of the functional equation that can be extended to the whole real axis are the linear functions and those containing $\arctan$.

Proof: A small calculation shows that the two new families are indeed solutions of the functional equation. For the second one, it suffices to do it for positive $b$. Here we use an addition formula for $\arctan$: $$\arctan(u)+\arctan(v)=\arctan(\tfrac{u+v}{1-uv})+\pi,$$ if $u,v\geq0$ and $uv>1$. This allows to show the functional equation for sufficiently large $x,y$; for all $x,y$ it follows because both sides of the functional equation are real analytic functions of $x,y$.

Another calculation shows that these new families consist of reciprocally convex/concave functions depending on the signs of $a$ and $b-c$ for the first one and of the sign of $a$ for the second one in the case $bc\geq0.$ If $bc<0$, then $\arctan(bx+c)$ is not concave as its derivative $\dfrac b{(bx+c)^2+1}$ is not monotonous on $(0,\infty)$.

It is much harder to show that these are all solutions. Actually, I did not guess the new family of solutions, but found them this "hard way".

Consider a differentiable solution $f$ of the functional equation such that $f'(x)\neq0$. Then, by Darboux's Theorem, $f'$ does not change sign, hence $f$ is strictly monotonous and its inverse function exists.

Claim 1: $f$ is infinitely often differentiable.

This follows from the functional equation which can be written $$\frac{xf'(x)+yf'(y)}{f'(x)+f'(y)}=f^{-1}(f(x)+f(y)-f(\tfrac{x+y}2)).$$ We fix $y$. Then the right hand side is a differentiable function of $x$. We name it $q(x)$. Then also $f'(x)=(y-q(x))f'(y)/(q(x)-x)$ is differentiable for $x\neq y$. Here we use that $x<q(x)<y$ if $x<y$ and $x>q(x)>y$ if $x>y$. Using several $y$, we see that $f'(x)$ is differentiable for all $x$ and hence $f$ is twice differentiable. This reasoning can be repeated and yields that $f$ is infinitely often differentiable.

Claim 2: $f$ satisfies the differential equation $\tag{1}f^{(4)}f'^2-6f^{(3)}f''f'+6f''^3=0.$

To show this, we replace $x$ by $x+h$ and $y$ by $x-h$ in the functional equation and obtain $$f(x+h)+f(x-h)=f(x)+f\left(x+h\frac{f'(x+h)-f'(x-h)}{f'(x+h)+f'(x-h)}\right).$$ Now we write down the Taylor expansion of $f(x+h)$ $$f(x+h)=a+bh+\frac12ch^2+\frac16dh^3+\frac1{24}eh^4+O(h^5)$$ with $a=f(x)$, $b=f'(x)$, etc. and expand both sides of the above equation up to $h^4$. This gives the differential equation.

It is not surprising to obtain a fourth order equation as a family of solutions contains up to four parameters: With $f(x)$ also $af(bx+c)+d$ are solutions. It was surprising for me, however, that this third order ODE for $f'$ can be explicitly solved. In the sequel,I present my original solution. A simpler solution can be found in Edit 2 at the end of this post.

To solve (1), we first introduce $d=\dfrac{f''}{f'}$. As then $d'=\dfrac{f^{(3)}}{f'}-\dfrac{f''^2}{f'^2}$ and $d''= \dfrac{f^{(4)}}{f'}-3\dfrac{f^{(3)}f''}{f'^2}+2\dfrac{f''^3}{f'^3}$, we obtain a second order ODE for $d$: $$\tag{2}d''-3d'd+d^3=0.$$ This equation is actually autonomous. An evident solution is $d=0$. It is the only constant solution. This leads to the family $f(x)=ax+b$.

Now we want to find the non-constant solutions of (2). On intervals on which $d$ is strictly monotonous, we introduce the composition $g=d'\circ d^{-1}$ of $d'$ and the inverse function of $d$. Then $d'(x)=g(d(x))$ and $d''(x)=g'(d(x))g(d(x))$. Replacing $d(x)$ by $s$, this gives an ODE for $g$ $$\tag{3}g'g-3sg+s^3=0.$$

It can be simplified introducing $h$ such that $g(s)=s^2h(s)$. This leads to the following ODE
for $h$ which turns out to be separable. $$\tag{4}shh'+2h^2-3h+1=0.$$

This equation has the constant solutions $h=1$ and $h=1/2.$ They lead to $g=s^2$ and $g=\frac12 s^2$, respectively, and thus to $d=-\frac 1{x+c}$ and $d=-\frac{2}{x+c}$ and finally to $f=a\ln(x+c)+b$ and $f=\frac a{x+c}+b$, respectively, which are solutions already mentioned in the question. Solving the separable equation (4) in the classical way (details omitted) for the general solution leads to $\frac{(h-1)^2}{2h-1}=\frac C{s^2}$ with a certain constant $C$. The equation for $g$ is then $\tag{5}g^2-2(s^2+C)g+s^2(s^2+C)=0.$ As its discriminant is $4C(s^2+C)$, we continue with positive $C$ and write $C=c^2$ with positive $c$. In the case of negative $C$, we could proceed differently, but it is simpler to consider the final result with purely imaginary $c$. So we find $$g(s)=s^2+c^2\pm c\sqrt{s^2+c^2}.$$

Next, we need to solve the separable equation $$\tag{6}d'=g(d)=d^2+c^2\pm c\sqrt{d^2+c^2}.$$ This is actually possible, because $$\int\frac{ds}{s^2+ c^2\pm c\sqrt{s^2+c^2}}=\frac{-2}{s+c\pm\sqrt{s^2+c^2}}+\mbox{ constant}$$ as can be verified by differentiation. The result is that only the negative sign is possible in (5) and that, neglecting the constant of integration for a moment, $$\frac{f''}{f'}=d=-\frac{2cx+2}{cx^2+2x}=-\frac c{cx+2}-\frac1{x}.$$ The general solution $d$ is obtained by replacing $x$ by $x+b$ with the constant of integration $b$.

This last linear first order equation is solved by $f'(x)=\dfrac{2a}{x(cx+2)}=\dfrac{a}{x}-\dfrac{ac}{cx+2}$ with another constant $a$. This finally leads to $$\tag{7}f(x)=a\ln\left(\frac{x+b}{c(x+b)+2}\right)+d,$$ where we reintroduced the constant of integration $b$. Rewriting this expression a bit and renaming the parameters $a,b,c,d$ we arrive at the first new family mentioned in the beginning.

In the case of negative $C$ in (5), we just consider purely imaginary $c=i\tilde c$ in (7). Rewriting this equation a bit and using the formula $$\arctan(z)=\frac1{2i}\log\left(\frac{1+iz}{1-iz}\right),$$ we arrive for certain (maybe complex) values of $a,b,c,d$ in (7) at $$f(x)=\tilde a\arctan(\tilde c x+\tilde b)+\tilde d$$ where $\tilde a,\tilde b,\tilde c,\tilde d$ are real. This finally proves the statement that there are only the solutions of the functional equation mentioned in the beginning of the answer.

Edit: There is a small gap in the proof. The function $g=d'\circ d^{-1}$ is defined on $d$-images of intervals on which $d$ is strictly monotonous. It might therefore be possible that solutions $d$ or $f$ are expressed by different formulas on different subintervals. This can, however, be excluded. As a solution of the differential equation (1) and as $f'(x)\neq0$, $f$ is actually real analytic. Therefore the identity theorem for analytic functions implies that the equations $f(x)=ax+b$ or ... are valid throughout the interval of definition of $f$. This closes the gap.

Edit 2: There is a simpler way to solve the 4-th order ODE (1), $f'(x)\neq0$, but it was not so obvious.

Put $q=1/f'$. Then $$q'=-\frac{f''}{f'^2},\ \ q''=2\frac{f''^2}{f'^3}-\frac{f^{(3)}}{f'^2},\ \ q^{(3)}=-6\frac{f''^3}{f'^4}+6\frac{f^{(3)}f''}{f'^3}-\frac{f^{(4)}}{f'^2}=0.$$ Hence $q$ is a non-zero polynomial of degree at most 2 and $$f'(x)=\frac1{Ax^2+Bx+C},$$ where $A,B,C$ are constants and do not vanish simultaniously.

If $A=B=0,C\neq0$ then we find $f(x)=\frac1Cx+E$ with some constant $E$.

If $A=0,B\neq0$ then we find $f(x)=\frac1B\ln(x+\frac CB)+E$ and $C/B$ has to be positive since $f$ is defined on $(0,\infty)$.

If $A\neq0$ then we must distinguish 3 cases for the discriminant $D=B^2-4AC$.

1. If $D>0$ then we can factor $Ax^2+Bx+C=A(x+b)(x+c)$ with real $b,c$. These have to be positive since $f$ is defined on $(0,\infty)$. Integrating we find $$f(x)=a\ln\left(\frac{x+b}{x+c}\right)+E$$ with certain $a,E$.

2. If $D=0$ then we can write $Ax^2+Bx+C=A(x+b)^2$ with $b=\frac B{2A}$. Again $b$ has to be positive. We obtain $$f(x)=\frac{a}{x+b}+E.$$

3. If $D<0$, $D=-F^2$ with positive $F$ then we can write $Ax^2+Bx+C=A(x+\frac B{2A})^2+\frac{F^2}{4A}$. We obtain $$f(x)=a\arctan(bx+c)+d$$ with certain $a,b,c,d$.

Since $f$ is real analytic as a solution of (1) with $f'(x)\neq0$, these expressions for $f$ are valid throughout $(0,\infty)$ by the identity theorem for analytic functions.