# Superhomogeneity of subadditive functions

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

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.
• Agree. Thanks. What happens if I also require $f$ to be continuous? Sep 26, 2021 at 23:22
• @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. Sep 26, 2021 at 23:41