Functions that satisfy $kf(x) \leq f(x/k)$ for $k \in (0,1]$ Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function  satisfying 
$$kf(x) \leq f(x/k)$$
for all $x \in \mathbb{R}$ and $k \in (0,1]$. Are there any standard results on what must necessarily be true about $f$? 
So far I have only been able to find some examples such as $f(x)=x^2$, and counterexamples such as $f(x)=e^x$.
 A: There are many such functions. Let $g$ be any positive increasing continuous function on $\mathbb R^{+}$. Then $f(x)=g(|x|)$ satisfies this inequality since $kf(x)\leq f(x)=g(|x|) \leq g(|\frac  x k|)=f(\frac x k)$. 
One special case: $g(x)=x^{\alpha}$ with $\alpha >0$. So any linear combination of the functions $|x|^{\alpha}$ with non-negative coefficients is also a solution. 
One necessary condition is $f \geq 0$ even if we know the inequality for one fixed $k<1$ : $f(x) \geq kf(kx)$ so $f(x) \geq k^{n}f(k^{n}x) \to (0)(f(0))=0$. 
Another necessary condition is $xf(x)$ must be increasing on $(0,\infty)$. To prove this let $0<x<y$. Then $x=ky$ where $k=\frac x y \in (0,1)$. Hence $kf(x)=kf(ky)\leq f(y)$ which means $\frac x y f(x) \leq f(y)$ or $xf(x) \leq yf(y)$.  This condition is also sufficient on $(0,\infty)$: if $xf(x)$ is increasing on $(0, \infty)$ the the given inequality holds for $x>0$ and $0<k \leq 1$
A: Any polynomial $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ such that $a_i \geq0$ satisfies $kf(x) \leq f(\frac{x}{k})$ for $k\in (0,1]$.
Note that since $k\in (0,1]$, $\frac{1}{k} \geq k$. Furthermore, since $k>0$, $\frac{1}{k}\geq k \Rightarrow\frac{1}{k^2} \geq k^2\Rightarrow\frac{1}{k^3} \geq k^3$, etc. 
$kf(x)=ka_0+ka_1x+ka_2x^2+\cdots+ka_nx^n\leq a_0+\frac{1}{k}a_1x+\frac{1}{k^2}a_2x^2+\cdots+\frac{1}{k^n}a_nx^n=f\left(\frac{x}{k}\right)$
Note that since $k\in(0,1]$, $a_0 \geq ka_0$.
