My proof uses derivatives. Can I extend it to non-differentiable functions? For every function $f:\mathbb{R}^+\to\mathbb{R}$ and positive constants $a,b$, define a function $f_{a,b}:(0,1)\to\mathbb{R}$:
$$ f_{a,b}(x) := f(2a\cdot  x) + f(2b\cdot(1-x))$$ 
A function $f$ is good if the function $f_{a,b}$ has a unique maximum at $x=1/2$ for every $a$ and $b$.
What are the good functions?
One family of good functions is the family of logarithmic functions,  $f(x)=c\cdot \ln{x}+d$, for some constants $c>0$ and $d$.  PROOF: If $f$ is logarithmic then $a$ and $b$ do not affect the maximization, so the maximum-point of $f_{a,b}$ is just the maximum point of the function: $c\ln{x} + c\ln{(1-x)}$. By standard calculus, the unique maximum is at $x=1/2$. 
My conjecture is that only the logarithmic functions are good. Currently I can prove this uniqueness only in the family of differentiable functions. PROOF: If $f$ is differentiable then:
$$ f_{a,b}'(x) := 2a\cdot f'(2a x) - 2b\cdot f(2b (1-x))$$ 
$$ f_{a,b}'(1/2) := 2a\cdot f'(a) - 2b\cdot f(b)$$ 
If 1/2 is a unique maximum point then $f_{a,b}'(x)=0$. Hence, for all $a,b$:
$$ a f'(a) = b f'(b)$$
so there exist constants $c,d$ such that:
$$ x f'(x) \equiv c$$
$$ f'(x) = c/x$$
$$ f(x) = c \cdot \ln x + d$$
so, if $f$ is differentiable and good, it must be a logarithmic function.
So, there are two options:


*

*There are other good functions, which are not differentiable.

*There are no other good functions.
I believe that option 2 is correct, but to prove it, the uniqueness proof from above should be modified to work without derivatives. Is this possible?

Claim. The functions $f(x):=c\log x +d$ with $c>0$ and $d\in{\mathbb R}$ are the only good functions.
Proof. The given condition on $f$ is equivalent with
$$f\bigl((1+t)a\bigr)+f\bigl((1-t)b\bigr)<f(a)+f(b)\qquad(a>0, \  b>0,\ 0<|t|<1)\ .\tag{1}$$
Letting $a:=(1-t)b$ here leads to $f\bigl((1-t^2)b)<f(b)$, hence $f$ is strictly increasing. From Lebesgue's theorem it then follows that $f$ is differentiable almost everywhere. Replacing $x$ with $x':=\lambda x$, if necessary,  we may assume that $f'(1)=:c\geq0$ exists.
If we put $t:={h\over a}$ and $b:=1$ in $(1)$ we obtain
$$f(a+h)-f(a)<f(1)-f\left(1-{h\over a}\right)\ ,$$
so that
$$\limsup_{h\to0}{f(a+h)-f(a)\over h}\leq \limsup_{h\to0}{f(1)-f\left(1-{h\over a}\right)\over h}={1\over a}f'(1)\ .$$
On the other hand, replacing $a$ by ${a\over 1+t}$ in $(1)$ we get
$$f\left({a\over1+t}\right)-f(a)>f\bigl((1-t)b\bigr)-f(b)\ .\tag{2}$$
If we now let $t:={-h\over a+h}$ and $b=1$ in $(2)$ then ${a\over 1+t}=a+h$ leads to
$$f(a+h)-f(a)>f\left(1+{h\over a+h}\right)-f(1)\ ,$$
so that we get
$$\liminf_{h\to0}{f(a+h)-f(a)\over h}\geq{1\over a}f'(1)\ .$$
Since $a>0$ was arbitrary we have proven that $f'(x)={c\over x}$ for all $x>0$ and some $c\in{\mathbb R}$. Since $f$ is strictly increasing we necessarily have $c>0$.
