My proof uses derivatives. Can I extend it to non-differentiable functions?

Question

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:

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

2. 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?

Answer 1

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 $c>0$.

Answer 2

Some incomplete thoughts:

Lemma 1. Let $f$ be good, $\alpha\in\Bbb R^+$, $\beta\in \Bbb R$. Then the function given by $g(x)= f(\alpha x)+\beta$ is also good.

Proof. Just note that $g_{a,b}(x)=f_{\alpha a,\alpha b}(x)+2\beta$. $\square$

Let $f$ be a good function.

Lemma 2. $f$ is strictly increasing.

Proof. Let $0<a<b<2a$. Then $f_{a,b}(\tfrac12)>f_{a,b}(\frac b{2a})$, i.e., $$ f(a)+f(b)> f(b)+f(2b(1-\tfrac b{2a}))=f(b)+f(a-\tfrac{(b-a)^2}{a}).$$ As we let $b$ vary over $(a,2a)$ we thus see that $f(a)>f(x)$ for $0<x<a$. $\square$

In particular, $\lim_{x\to 0^+}f(2b(1-x))$ exists and is $\le f(2b)$ for every $b>0$.

Assume $\inf f=L>-\infty$. Then $\lim_{x\to 0^+}f(2ax)=L$ for all $a$. We conclude $$ f(a)+f(b)\ge L+\lim_{x\to 2b^-}f(x)$$ and $a$ can be picked with $f(a)\approx L$, we arrive at $f(b)\ge \lim_{x\to 2b^-}f(x)\ge f(\frac32b) $, contradicting lemma 2. We conclude that $$\inf f=-\infty. $$

Define $g(x)=f(\tfrac 1x)$. Then $f_{1/a,1/b}(\tfrac12)>f_{a,b}(\tfrac{a}{a+b})$ translates into $$\frac{g(a)+g(b)}2>g\Bigl(\frac{a+b}2\Bigr).$$ As $g$ is strictly decreasing, we conclude that $g$ is left-continuous, i.e., $f$ is right-continuous, $f(x)=\lim_{h\to 0^+}f(x+h)$.

Fix $\theta$ with $0<\theta < 2$. Then $$f(a)-f(\theta a)>f(b(1- \tfrac \theta 2))-f(b),$$ i.e., $x\mapsto f(x)-f(\theta x)$ is bounded from below. Then of course also $x\mapsto f(x)-f(\theta x)+f(\theta x)-f(\theta^2 x)=f(x)-f(\theta^2 x)$ is bounded from below. We conclude that $x\mapsto f(x)-f(\theta x)$ is bounded from below for all $\theta >0$. Thus let $$\ell(\theta)=\sup\{\,f(\theta x)-f( x)\mid x>0\,\}.$$ Then $\ell(\theta_1\theta_2)\le \ell(\theta_1)+\ell(\theta_2)$.