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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 $\left(0,1\right]$.

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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 $f(tx)=1<tf(x)$.

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  • $\begingroup$ Agree. Thanks. What happens if I also require $f$ to be continuous? $\endgroup$ Sep 26, 2021 at 23:22
  • $\begingroup$ @CharlesPehlivanian : Perhaps one can modify/smooth out the $f$ in this example to make it continuous, while preserving the other properties. However, at this point I don't have a clear idea how to do that. You might want to ask the "continuous" version of your question separately. $\endgroup$ Sep 26, 2021 at 23:41

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