# When is a continuous subadditive function (0,1]-superhomogeneous

Continuous version of this Superhomogeneity of subadditive functions

Let $$f$$ be a continuous 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 $$\left(0,1\right]$$.


Indeed, suppose that we have a function $$f\colon[0,\infty)\to\R$$ such that $$f(0)=0$$ and $$f$$ is continuous, nonnegative, nondecreasing, and subadditive, but not superhomogeneous.

Then, for $$\C:=[0,\infty)^n$$, the function $$h\colon\C\to\R$$ given by the formula $$h(x):=f(|x|)$$ for all $$x\in\C$$ (where $$|x|$$ is the Eucludean norm of $$x$$) will be such that $$h(0)=0$$ and $$h$$ is continuous, nonnegative, nondecreasing in each of the coordinates of the argument, and subadditive, but not superhomogeneous on $$\C$$. In particular, the subadditivity of $$h$$ will follow from the subadditivity and monotonicity of $$f$$ and the triangle inequality $$|x+y|\le|x|+|y|$$ for $$x$$ and $$y$$ in $$\C$$.

So, it is enough to construct a function $$f$$ as above. To do that, let $$\begin{equation*} f(x):=f_m(x):=m+g(x-(2^m-1))\text{ if }x\in J_m:=[2^m-1,2^{m+1}-1) \end{equation*}$$ for some $$m=0,1,\dots$$, where $$\begin{equation*} g(u):=\min(1,2u). \end{equation*}$$ Then $$f(0)=0$$ and $$f$$ is continuous, nonnegative, nondecreasing, and subadditive, but not superhomogeneous on $$\C$$.

Here is the graph $$\{(x,f(x))\colon0\le x\le2^5\}$$:

Details: It is easy to see that $$f(0)=0$$ and $$f$$ is continuous, nonnegative, and nondecreasing. Also, if e.g. $$x\in\mathcal C$$, $$1, and $$t=1/x$$, then $$t\in(0,1)$$, $$f(x)=f_1(x)=1+2(x-1)$$, $$f(tx)=f_1(tx)=1$$, and hence $$f(tx), so that $$f$$ is not superhomogeneous on $$\C$$.

It remains to check that $$f$$ is subadditive. Take any $$x$$ and $$y$$ in $$[0,\infty)$$. Without loss of generality $$x\le y$$, so that $$x\in J_k$$ and $$y\in J_m$$ for some nonnegative integers $$k$$ and $$m$$ such that $$k\le m$$. So, $$x+y<2^{k+1}-1+2^{m+1}-1<2^{m+2}-1$$. Also, the function $$g\colon[0,\infty)\to\R$$ is nondecreasing and concave, with $$g(0)=0$$, so that $$g$$ is subadditive. So, if $$k\ge1$$, then \begin{equation*} \begin{aligned} f(x+y)&\le f_{m+1}(x+y) \\ &=m+1+g(x+y-(2^{m+1}-1)) \\ &\le k+m+g(x+y-(2^{m+1}-1)) \\ &\le k+m+g(x-(2^k-1)+y-(2^m-1)) \\ &\le k+m+g(x-(2^k-1))+g(y-(2^m-1)) \\ &=f_k(x)+f_m(y)=f(x)+f(y). \end{aligned} \tag{0} \end{equation*}

If $$k=m=0$$, then $$x$$ and $$y$$ are in $$J_0=[0,1)$$, and
$$f(x)+f(y)=g(x)+g(y)$$. So, if $$x+y<1$$, then $$f(x+y)=f_0(x+y)\le g(x+y)$$, and the inequality $$f(x+y)\le f(x)+f(y)$$ follows because $$g$$ is subadditive. If now $$x+y\ge1$$, then $$x+y\in J_1$$ and hence $$f(x+y)=f_1(x+y)=1+g(x+y-1)=g(1)+g(x+y-1)\le g(x)+g(y)$$ by the concavity of $$g$$. So, $$f(x+y)\le g(x)+g(y)=f(x)+f(y)$$.

It remains to consider the case $$k=0, so that $$x\in J_0=[0,1)$$, $$f(x)=g(x)$$, and $$\begin{equation*} f(x)+f(y)=g(x)+m+g(y-(2^m-1)). \tag{1} \end{equation*}$$ If now $$x+y<2^{m+1}-1$$, then $$\begin{equation*} f(x+y)\le f_m(x+y)=m+g(x+y-(2^m-1)) \\ \le m+g(x)+g(y-(2^m-1)), \end{equation*}$$ since $$g$$ is subadditive. So, by (1), $$f(x+y)\le f(x)+f(y)$$.

It remains to consider the case when $$k=0 and $$x+y\ge2^{m+1}-1$$. Then $$y\ge2^{m+1}-1-x\ge2^{m+1}-2\ge2^m$$, since $$m\ge1$$. Therefore and because $$g$$ is nondecreasing, $$\begin{equation*} f(x)+f(y)=g(x)+m+g(y-(2^m-1)) \\ \ge g(x)+m+g(1)=g(x)+m+1. \tag{2} \end{equation*}$$ As in multiline display (0), $$\begin{equation*} f(x+y)\le m+1+g(x+y-(2^{m+1}-1)). \end{equation*}$$ Therefore and because $$y\in J_m$$ and $$g$$ is nondecreasing, we have $$\begin{equation*} f(x+y)\le m+1+g(x+y-(2^{m+1}-1)) \\ \le g(x)+m+1\le f(x)+f(y), \end{equation*}$$ by (2). This completes the proof.

Moreover, we can construct a function $$\tf\colon[0,\infty)\to\R$$ such that $$\tf(0)=0$$, $$\tf(x)=2x$$ in a right neighborhood of $$0$$, $$\tf$$ is infinitely smooth on $$(0,\infty)$$, and still is nonnegative, nondecreasing, and subadditive, but not superhomogeneous. So, the function $$\tih\colon\C\to\R$$ given by the formula $$\tih(x):=\tf(|x|)$$ for all $$x\in\C$$ will be such that $$\tih(0)=0$$, $$\tih(x)=2|x|$$ in a neighborhood of $$0$$ in $$\C$$, $$\tih$$ is continuous on $$\C$$, infinitely smooth on the interior of $$\C$$, nonnegative, nondecreasing in each of the coordinates of the argument, and subadditive, but not superhomogeneous on $$\C$$.

Indeed, let $$K\colon(0,\infty)\to\R$$ be any nonnegative infinitely smooth function supported on a small neighborhood, say $$V$$, of $$1$$ and such that $$\int_0^\infty cK(c)\,dc=1$$. Let then the function $$\tf\colon[0,\infty)\to\R$$ be defined as the multiplicative convolution with $$K$$ of the function $$f$$ constructed above: $$$$\tf(x):=\int_0^\infty f(cx)K(c)\,dc$$$$ for $$x\in[0,\infty)$$. So, the function $$\tf$$ is a mixture of the horizontal rescalings $$[0,\infty)\ni x\mapsto f(cx)$$ of $$f$$. So, $$\tf(0)=0$$ and $$\tf$$ is nonnegative, nondecreasing, and subadditive.

Moreover, if the neighborhood $$V$$ of $$1$$ is small enough, then $$\tf$$ will be close enough to $$f$$ and thus not superhomogeneous. Furthermore, if the neighborhood of $$1$$ is small enough, then for all $$x$$ in some right neighborhood of $$0$$ we will have $$$$\tf(x)=\int_0^\infty 2cx K(c)\,dc=2x.$$$$ Finally, for any $$x\in(0,\infty)$$, $$$$\tf(x)=\frac1x\,\int_0^\infty f(u)K\Big(\frac ux\Big)\,du,$$$$ which shows that $$\tf$$ is infinitely smooth on $$(0,\infty)$$.

Here is the graph $$\{(x,\tf(x))\colon0\le x\le2^3\}$$ for $$K(c):=\frac1C\,\exp\Big(-\frac1{1/4-(c-1)^2}\Big)\,1(|c-1|<1/2)$$ and $$C:=\displaystyle{\int_{1/2}^{3/2} \exp\Big(-\dfrac1{1/4-(c-1)^2}\Big)\,dc}$$:

• Comprehensive answer. It is easy to show that convexity in addition to subadditivity gives us the superhomogeneity condition, I was wondering if that could be weakened at all. Not in this way at least. Thanks. Sep 28, 2021 at 12:14