Are functions satisfying a certain inequality monotone

Question

Let $f: (0, \infty) \rightarrow (0,\infty)$ be a continuous function which satisfies the inequality $$f(x) + f(y) \geq 2f(x+y).$$ Is $f$ necessarily monotone?

Answer 1

To construct a counterexample, let us start from the function $f(t)=e^{-t}$, which satisfies the main inequality ($f(x)+f(y)\geq 2f(x+y)$ for all $x,y>0$): $$e^x+e^y\geq 2 \Rightarrow e^{-x}+e^{-y}\geq 2e^{-x-y},\qquad \forall x,y>0 $$ Now let $a>0$. We want to redefine $f$ (only) on $[a,+\infty)$ so that $f$ is still continuous and it satisfies the main inequality, while not being necessarily monotonic. To this goal we add the following requirement for $f$: $$ e^{-t}\leq f(t)\leq \beta e^{-t},\qquad \forall t\geq a\qquad (*) $$ for some $\beta >1$. This is not too restrictive as there clearly are plenty of non-monotonic functions satisfying these requirements. On the other hand, we now verify that such a function always satisfies the main inequality.

Pick $x,y>0$ with $x\leq y$. When $x+y\leq a$ (which implies $x,y\leq a$) the inequality blatantly holds because it does for the function $e^{-t}$ (recall that $f(t)=e^{-t}$ if $t\leq a$). That leaves only the case where $x+y\geq a$.

If $x+y\geq a$, from $(*)$ we get $$f(x)+f(y)\geq e^{-x}+e^{-y}$$ $$2f(x+y)\leq 2\beta e^{-x-y} $$ Thus the main inequality holds if $$e^{x}+e^y\geq 2\beta$$ But $x+y\geq a$, and since the function $x\mapsto e^x+e^{a-x}$ has a minimum at $x=a/2$, we get $$e^x+e^y\geq e^x+e^{a-x}\geq 2e^{a/2} $$ and $$2e^{a/2}\geq 2\beta \iff \beta \leq e^{a/2} $$ which is reasonable since $e^{a/2}>1$ when $a>0$.

In conclusion if $1<\beta\leq e^{a/2}$ then a function $f$ with the properties described above satisfies the main inequality. Yet, such a function is still not necessarily monotonic.

Comment: I find it interesting how a function $f$ satisfying ( * ) need only be equal to $e^{-t}$ on an (arbitrary small) interval $[0,a]$, whereas it can take different values for all $t\geq a$. On the other hand we cannot let it take arbitrary values for all $t>0$ (i.e. setting $a=0$), otherwise it might not satisfy the main inequality, regardless of the value of $\beta$. Indeed, the main inequality tells us that $f(x)\leq f(0)$ for all $x$ (choose $y=0$), and if $a=0$ this condition may not be fulfilled. Instead, if $a>0$, we have $f(x)\leq \beta e^{-x}\leq \max \left\{\beta e^{-a};f(0)\right\}$ and $\beta e^{-a}\leq \beta e^{-a/2}\leq 1=f(0)$.