A beautiful inequality for convex functions

Solution 1:

The statement holds for any increasing $f:[0,1]\to[0,+\infty)$, i.e. the continuity of $f$ is not necessary. Note that if the statement holds for some $f$, then after adding a common positive constant to $f,g,h$ simultaneously, the statement still holds, so for simplicity, we may assume $f(0)=0$.

Following your notations, define $H$ by letting $H(x)= 1$ when $x\ge 0$ and $H(x)=0$ when $x<0$. Define $V(x):=xH(x)$. Let us start with the simplest case $f(x)=f_t(x):=H(x-t)$ for some $t\in (0,1)$ and construct the associated $g_t$ and $h_t$. Functions of the form $x\mapsto b\cdot V(x-a)$ for appropriate constants $a< 1$ and $b>0$ are natural candidates for both $g_t$ and $h_t$, which are non-negative, continuous, increasing and convex. Using the constraints $g_t\le f_t\le h_t$ and $\frac{1}{2}\int_0^1 h_t\le\int_0^1 f_t\le 2\int_0^1 g_t$, we can easily find $g_t$ and $h_t$ with the supposed form as follows:

$$g_t(x):= \frac{1}{1-t}V(x-t),\quad h_t(x):= \frac{1}{1-t}V(x-2t+1).\tag{1}$$ It is easy to verify that $g_t$ and $h_t$ satisfy all the requirements in the statement for $f_t$.

When $f$ is an increasing $C^1$ function with $f(0)=0$, it can be expressed as $$f(x)=\int_0^1 f_t(x)f'(t)dt.$$ As a result, $g$ and $h$ can be constructed by expressing them in a similar way in terms of $g_t$ and $h_t$ respectively:

$$g(x):=\int_0^1 g_t(x)f'(t)dt,\quad h(x):=\int_0^1 h_t(x)f'(t)dt.\tag{2}$$ Due to $(2)$ and the fact $f'\ge 0$, it is easy to check that all the requirements for $g$ and $h$ are inherited from that for $g_t$ and $h_t$.

When $f$ is only increasing, the correct formulas of $g$ and $h$ can be guessed from $(1)$ and $(2)$ using integration by parts. Simple calculation shows that when $x\in[0,1)$, $$g(x)=(1-x)\int_0^x\frac{f(t)}{(1-t)^2}dt, \quad h(x)=(1-x)\int_0^{\frac{1+x}{2}} \frac{f(t)}{(1-t)^2}dt.\tag{3}$$

It is easy to check that $$g(1):=\lim_{x\to 1^-}g(x)=\lim_{x\to 1^-}f(x),\quad h(1):=\lim_{x\to 1^-}h(x)=2\lim_{x\to 1^-}f(x),\tag{4}$$ and $g$ and $h$ defined in $(3)$ and $(4)$ satisfy all the requirements.