Let $f : \mathbb{R}^n \to \mathbb{R}_+$ satisfy $$f(a) - f(b) \le C f(a - b)$$ $\forall a, b \in \mathbb{R}^n$ for some $C \ge 0$. Is there a name for such functions? (I would be happy to have a name for the special case $C=1$.)

Note the similarity to $$\|x\| - \|y\| \le \|x - y\|$$ which is satisfied by all norms.

  • 1
    $\begingroup$ Just a small note: if $C < 1$, then plugging in $a=x$ and $b=0$ yields that $f(x)$ is bounded above by $f(0)/(1-C)$, and if $C>1$ then $f$ is bounded below by this quantity. $\endgroup$ Nov 12 '20 at 17:01
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    $\begingroup$ Relabel $a = c + b$ then your inequality reads f(c+b) <= C f(c) + f(b). When C = 1 this is "subadditive" $\endgroup$ Nov 12 '20 at 17:29
  • $\begingroup$ There's a name in the case $C = 0$. :-) $\endgroup$
    – LSpice
    Nov 12 '20 at 17:35
  • $\begingroup$ @StevenGubkin Along similar lines, $a=2x$ and $b=x$ leads to $f(2^n x) \le (1 + C)^n f(x)$. $\endgroup$
    – user27182
    Nov 12 '20 at 18:10
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    $\begingroup$ For, $h \rightarrow 0$, and any $x$, $f'(x)=\frac{f(x+h)-f(x)}{h} \leq Cf'(h)/h=T$. Here, we take $f(0)=0$ and $f(h)/h=f'(0)=T$. and If, $C<1$, $f(x)-f(0)=f(x)\leq Cf(x)$ implies that $f(x)\leq 0$ for all $x$. And this two can't happen simultaneously. So, $C$ can't be less than $1$, except for zero function. $\endgroup$
    – Alapan Das
    Nov 13 '20 at 4:02

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