# Terminology for $f(tx) \leq tf(x)$ when $0\leq t \leq 1$

That $$f(tx) \leq tf(x)$$ for $$0\leq t \leq 1$$ is a famous property of convex functions, often used, for example, in the proof that convex functions are superadditive on the positive reals.

Edit: "A famous property of convex functions such that $$f(0) \geq 0$$", as the Wikipedia article mentions, sorry for not including that extra condition in my original post.

In economics, properties like $$[f(tx) \le tf(x)$$ for $$0\le t \le 1]$$ are often discussed in terms of "returns to scale".

My question is, in mathematics, is there a common name for the property $$[f(tx) \le tf(x)$$ for $$0\le t \le 1]$$?

• Remark: This is like a weaker version of convexity where the only secant lines that are required to lie above the graph are the ones through $(0,0)$. Feb 25, 2022 at 17:47

It is called a starshaped function, see definition 2 in [1].

Reference:

[1] A. M. Bruckner and E. Ostrow, "Some Function Classes Related to the Class of Convex Functions", 1962.

There are convex functions that do not satisfy this property:

The functions $$f(x)=x^2+a$$ satisfy the property $$\tag{1} f(tx)\le tf(x)\,\quad\text{ for }0\le t\le 1$$ only when $$a\le 0\,.$$ Indeed, $$t^2x^2+a\le tx^2+at$$ implies $$a\le 0$$ just by taking the case $$t=0\,.$$

In general it follows from (1) that $$f(0)\le tf(0)\,,\,\forall t\in[0,1]\,.$$ So that

• $$f(0)\le 0\,.$$

The property (1) is equivalent to $$\tag{2} \frac{f(x)}{x}\quad\text{ being increasing on } (-\infty,0)\text{ and on }(0,+\infty)\,.$$ Let $$0 Then $$t=y/x\in(0,1)$$ and (1) implies $$f(y)=f(tx)\le tf(x)=\frac{y}{x}f(x)\,.$$ Therefore $$f(x)/x$$ is increasing on $$(0,+\infty)\,.$$ Let $$-\infty Then $$t=x/y\in(0,1)$$ and (1) implies $$f(x)=f(ty)\le tf(y)=\frac{x}{y}f(y)\,.$$ Since $$x<0$$ this implies $$\frac{f(x)}{x}\color{red}{\ge}\frac{f(y)}{y}$$ so that $$f(x)/x$$ is also increasing on $$(-\infty,0)\,.$$ Conversely, assume that (2) holds. If $$x<0$$ then for $$0 the point $$y=tx$$ is in $$(x,0)\,.$$ Then $$\frac{f(x)}{x}\le\frac{f(y)}{y}=\frac{f(tx)}{tx}\,.$$ Because $$x<0$$ this implies $$f(tx)\le tf(x)\,.$$ If $$x>0$$ then $$y=tx$$ is in $$(0,x)$$ and $$\frac{f(y)}{y}=\frac{f(tx)}{tx}\le\frac{f(x)}{x}$$ which implies (1).

• Doesn’t answer the question, does it? The OP asked about the name of this property. Feb 26, 2022 at 12:05
• I doesn't. I am more interested in properties than in names. Esp. when a comment conjectured that is has something to do with convex functons. Feb 26, 2022 at 12:29

Such functions are said to be subhomogeneous. There’s a pattern here. If you have a name xxx for functions that satisfy an equality, then the name sub-xxx is used for functions that satisfy a similar inequality. So ….
Linear —> sublinear.
• @bubba . According to this link a function $f(x)$ is subhomogeneous when $f(tx)\le tf(x)$ for all $t>1$. A function is homogeneous (of degree one) when $f(tx)=tf(x)$ for all $t$. OP's property makes those requirements only for $0\le t\le 1$. Feb 27, 2022 at 5:54