Let $f$ be a function of $\geq 2$ real variables defined on a convex cone $\mathcal{C}$ in the upper half plane, with $f(0) = 0$. Suppose $f$ is subadditive, i.e. $f(x\_1+y\_1, \dots, 
x\_n+y\_n) \leq f(x\_1, \dots, x\_n) + f(y\_1, \dots, y\_n)$ in its domain, $f \geq 0$, and $f$ is nondecreasing in $x$ and nonincreasing in $y$.


I'd like to know whether $f$ is necessarily $\left(0,1\right]$-superhomogeneous, i.e., whether $f(\lambda x) \geq \lambda f(x)$ for $\lambda \in \left(0,1\right]$.


I can show that $f$ is superhomogeneous for $\lambda$ as above in $\mathbf{N}^{-1}$, I'd like to know if that's true on the interval $\left(0,1\right]$.



Superhomogeneity of subadditive functions
The answer is no. E.g., let $f(0):=0$ and $f(x):=1+m\_+$ if $|x|\in[2^{m-1},2^m)$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $m$; that is, $f(x)=1+(1+\lfloor\log\_2|x|\rfloor)\_+$ for all $x\in\mathcal C\setminus\{0\}$. Here, $m\_+:=\max(0,m)$ and $|x|$ is the Eucludean norm of $x$.


**Details:** Clearly, $f$ is nonnegative and nondecreasing in each of the coordinates of the argument. Take now any $x$ and $y$ in $\mathcal C$ such that $|x|\in[2^{k-1},2^k)$ and $|y|\in[2^{m-1},2^m)$ for integers $k$ and $m$ such that $k\le m$. Then $f(x)+f(y)=1+k\_+ +1+m\_+\ge2+m\_+$ and $|x+y|\le|x|+|y|<2^k+2^m\le2^{m+1}$, whence
$f(x+y)\le1+(m+1)\_+\le2+m\_+\le f(x)+f(y)$, so that $f$ is subadditive.


However, the condition $f(tx)\ge tf(x)$ will fail to hold if e.g. $x\in\mathcal C$, $|x|=1$, and $t\in(1/2,1)$ -- because then $f(x)=2$ whereas $f(tx)=1<tf(x)$.


