Analysis of the functional inequality $f(x-s)f(y+s) \geq f(x)f(y)$

Question

I would like to find references or analysis of the two functional inequalities:

1. $f(x-s)f(y+s)\geq f(x)f(y)$
2. $f(x-s)+f(y+s) \geq f(x) + f(y)$

where $x,y\in \mathbb{R}$, $y>=x$ and $s>0$. Essentially these functions satisfy that their "spread" products and sums are greater than the sums of products at the original $x$ and $y$. I have found similarities in Supermodularity and functions of increasing differences.

Positive exponential functions $f(x)=a*e^{bx}$ trivially satisfy the first inequality.

Secondly, any function which satisfies the first has its logarithm satisfying the second as $$ f(x-s)f(y+s)\ge f(x)f(y) \iff \ln f(x-s) + \ln f(y+s) \ge \ln f(x) +\ln f(y)$$

Affine functions $f(x)=ax+b$ do not satisfy (1) as it results in the inequality $$a^2s(x-y-s)\ge 0$$ which is only true when $a=0$ and the function is constant.

What can be said about such functions? Must they be convex?

Answer 1

Assuming continuity*, your second condition is equivalent to the condition that $f$ is convex, and the first condition is equivalent to $f$ having the form either $f\equiv 0$, $f= e^g$, or $f=-e^g$, for $g$ convex.

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Second condition

To see why the second condition is equivalent to convexity, consider $f$ satisfying the second condition, and note that for $a<b$, taking $s=\frac{b-a}{2}$ and $x=y=\frac{a+b}{2}$ we obtain

$$ f(a)+f(b)\geq 2f\left(\frac{a+b}{2}\right),$$ so that $$ f\left(\frac{a+b}{2}\right)\leq \frac{f(a)+f(b)}{2}, \tag{1}$$ one of many equivalent characterizations of convexity for continuous functions.

In the reverse direction, the easiest way to see that convexity implies your condition 2 is probably to recall that convexity is equivalent to the statement that for $a<b<c$ the segments joining $(a,f(a))$ to $(b,f(b))$ and joining $(b,f(b))$ to $(c,f(c))$ lie below (not necessarily strictly) the segment joining $(a,f(a))$ to $(c,f(c))$.

Applying this twice tells us that the segment $A$ joining $(x,f(x))$ to $(y,f(y))$ lies below the segment joining $(x,f(x))$ to $(y+s,f(y+s))$, and that segment in turn lies below the segment $B$ joining $(x-s,f(x-s))$ to $(y+s,f(y+s))$. The midpoint of $A$ is $(\frac{x+y}{2},\frac{f(x)+f(y)}{2})$, and the midpoint of $B$ is $(\frac{x+y}{2},\frac{f(x-s)+f(y+s)}{2})$, so since they have the same first coordinate, we obtain $$\frac{f(x-s)+f(y+s)}{2}\geq \frac{f(x)+f(y)}{2},$$ so that your condition is satisfied.

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First condition

Your first condition becomes equivalent to the condition that $f=e^g$ for $g$ convex, provided that $f$ is positive everywhere, as you have observed you can take the natural logarithm and recover the second condition. If $f$ is negative everywhere, then since $-f$ satisfies the same condition, we have $f=-e^g$ for convex $g$.

Finally, if $f$ is $0$ at even a single point, then if $(x-s,y+s)$ is an interval on which $f$ is either always positive or always negative, and $f(y+s)=0$ or $f(x-s)=0$, then we obtain a contradiction. Therefore since $\{f>0\}$ and $\{f<0\}$ are open sets, hence disjoint unions of open intervals, we see that $f$ vanishing at a single point must imply that $f\equiv 0$ everywhere.

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*Remark

I do not know whether the continuity assumption is necessary - my understanding is that weaker definitions of convexity, such as inequality (1), may allow pathological discontinuous functions constructed with the axiom of choice, though I don't have a reference for this fact. I also don't know if your characterization, which is a priori stronger, might be enough to recover continuity.