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Given $a<0$ such that $k:=2^{a}+3^{a}\in (0,1)$ and define $f(x):=x^{a}$ for all $x>0$. Then,

$f(2x)+f(3x)=x^{a}(2^{a}+3^{a})=kf(x)$.

Somebody know another example of a function $f:(0,+\infty)\longrightarrow [0,+\infty)$ (not identically null) such that

$f(2x)+f(3x)\leq k f(x)$,

for some $k\in (0,1)$ and $x>0$.

Thank you in advance for your comments!

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    $\begingroup$ If it's continuous at $x=0$ then taking $x\to 0$ gives $2\leq k$. Try to consider functions that have a divergence as $x\to 0$, for example $e^{-x}/x^n$. $\endgroup$ – Winther Apr 25 at 16:55
  • $\begingroup$ Thank you! Nice example! $\endgroup$ – user123043 Apr 26 at 17:17

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