5
$\begingroup$

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 $0\le t \le 1]$?

$\endgroup$
1
  • 2
    $\begingroup$ 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 $(0,0)$. $\endgroup$ Feb 25, 2022 at 17:47

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
2
0
$\begingroup$

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 $0<y<x<\infty\,.$ 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 $-\infty<y<x<0\,.$ 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 $0<t<1$ 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).

$\endgroup$
2
  • 2
    $\begingroup$ Doesn’t answer the question, does it? The OP asked about the name of this property. $\endgroup$
    – bubba
    Feb 26, 2022 at 12:05
  • $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Feb 26, 2022 at 12:29
0
$\begingroup$

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.
Additive —> subadditive.
Homogeneous —> subhomogeneous.
And so on.

I wouldn’t say the name subhomogeneous is common, but it’s not unknown, and it’s meaning is easy to guess.

$\endgroup$
3
  • 2
    $\begingroup$ Can you provide a reference to this usage of "subhomogeneous"? It's a sensible term but I wasn't able to find it used for this property. $\endgroup$ Feb 26, 2022 at 17:28
  • $\begingroup$ I just googled “subhomogeneous function”. There were numerous hits. Here is one: sciencedirect.com/science/article/pii/S0022247X13001972 $\endgroup$
    – bubba
    Feb 26, 2022 at 23:53
  • 1
    $\begingroup$ @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 $0\le t\le 1$. $\endgroup$
    – Kurt G.
    Feb 27, 2022 at 5:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .