Find the continuous functions with the property that there is an $a\in\mathbb{R^{+}}$ and a $k\in\mathbb{N}$ such that $f(x)f(2x)...f(nx)\leq an^{k}$.

Question

Find the continuous functions $f:\mathbb{R}\rightarrow[1,\infty)$ with the property that there is an $a\in\mathbb{R^{+}}$ and a $k\in\mathbb{N}$ such that $f(x)f(2x)...f(nx)\leq an^{k}$, for any $x\in\mathbb{R}$ and $n\in\mathbb{N^{*}}$. But to search for a function that fullfils this property is out of my sight. I suppose it's an exponential function in order to pass from product to sum and exploit the inequality properly. I tried to differentiate it but it's wrong from multiple reasons listed in the comments section, and I am thankful towards the sensible observations made there.

Answer 1

Fix any $y\in {\mathbb R}$, and let $x=y/n$.

Then

$$ \frac{1}{n} \sum_{j=1}^n \ln f (\frac{j y}{n}) \le \frac{1}{n} \left ( \ln a + k \ln n\right).$$

The LHS is a Riemann sum converging to $\int_0^y \ln f(t) dt$ as $n\to\infty$ if $y>0$ and to $\int_y^0 \ln f(t) dt$ if $y<0$. The RHS converges to $0$. From this we conclude that $f\equiv 1$.