Originally Posted by
Krizalid Let $\displaystyle g(x)=e^{-Kx}f(x),$ thus the only thing we need to prove is that $\displaystyle g(x)=0.$
We have that $\displaystyle g'(x)=e^{-Kx}\big(f'(x)-Kf(x)\big),$ but given $\displaystyle |f'|\le K|f|$ we get that $\displaystyle g(x)g'(x)\le0$ which is the same as $\displaystyle \left( \frac{g^{2}(x)}{2} \right)'\le 0,$ thus $\displaystyle \big(g(x)\big)^2\le0$ then $\displaystyle g(x)=0$ for $\displaystyle x\ge0.$
We proved a stronger result, given $\displaystyle f:[0,\infty)\to\mathbb R$ differentiable.