The following problem is only tangently related to my present work, and I do not have any applications. However, I am curious to know the solution -- or even to see a lack thereof, indicating that the problem may be worth a serious research.
Let $\mathcal F$ denote the class of all functions $f\colon[0,1]\to{\mathbb R}$ satisfying the inequality $$ f(tx+(1-t)y) \le tf(x) + (1-t)f(y) + |y-x|, \ x,y,t \in [0,1] $$ and the boundary condition $\max{f(0),f(1)}\le 0$.
Substituting $x=0$ and $y=1$, we see that all functions from $\mathcal F$ are
uniformly bounded from above by $1$, and we let
$$ F(x) := \sup \{ f(x)\colon f\in {\mathcal F} \}. $$
An interesting observation is that the function $F$ itself belongs to
$\mathcal F$; hence, it is the pointwise maximal function of this class. What
is this function, explicitly?
It is not difficult to see that $F$ is continuous on $[0,1]$, symmetric
around $x=1/2$, positive on $(0,1)$, and vanishing at $x=0$ and $x=1$.
Substituting $t=x$ and $y=0$ and renaming the variables in the resulting
estimate gives $F(x)\le 2\sqrt x$; hence, indeed,
$$ F(x) \le \min\{2\sqrt x,1,2\sqrt{1-x} \}. $$
On the other hand, the functions $\min\{4x,1,4(1-x)\}$ and $ex\ln(1/x)$
belong to $\mathcal F$, implying
$$ F(x) \ge \min\{4x,1,4(1-x)\} $$
and
$$ F(x) \ge \max \{ ex\ln(1/x), e(1-x)\ln(1/(1-x))\}, $$
for all $x\in[0,1]$.
The graphs of the bounding functions in the right-hand sides:
(Thus, the graph of $F$ resides somewhere between the highest of the green curves and the red curve.)
Comparing the estimates, we see that $F(x)=1$ for $x\in[1/4,3/4]$, and
$0<F(x)<1$ for $x\in(0,1/4)$ and also for $x\in(3/4,1)$. I have more
estimates and observations of this sort. Say, I can show that if $x=e^{-k}$,
then $F(x)=ex\ln(1/x)$; that is, $F(e^{-k})=ke^{1-k}$ for
$k\ge 2$. Another funny fact is that for any function $f\in{\mathcal F}$
(and in particular for the function $f=F$) one has
$$ f(tx+(1-t)y) \le tf(x) + (1-t)f(y) + F(t)|y-x|, \ x,y,t \in [0,1]. $$
However, with all these partial results and rather tight bounds, so far I was
unable to find $F(x)$ in general. Any ideas?
