Prove equality of functions satisfying a periodic relation

Question

I have two functions $f_1,f_2:[0,\infty)\to[0,1]$ such that:

1. $f_1(0)=f_2(0)=1$.
2. $f_1(\infty)=f_2(\infty)=0$.
3. $f_1$ and $f_2$ are non-increasing.
4. $f_1$ and $f_2$ are differentiable.
5. For a fixed $c>0$, $f_1(x+c)/f_1(x)=f_2(x+c)/f_2(x)$ for all $x\geq0$.

I want to prove that $f_1(x)=f_2(x)$ for all $x\geq0$ or to find a counterexample on this claim.

What I tried: The first and fifth points imply that $f_1(kb)=f_2(kb)$ for $k\in\mathbb{N}$. The counterexamples I am able to find involve discontinuous functions $f_1$ and $f_2$. Defining $g(x)=f_1(x)/f_2(x)$, then $g(x+b)=g(x)$ and I was able to apply the mean value theorem in $(0,b)$.

Answer 1

Counterexamples do exist, consider $$f_1(x)=e^{-\pi x/c},\ \ f_2(x)=f_1(x)\left(1+\sin^2\left(\frac{\pi x}c\right)\right).$$ Property 1, 2, 4 are easily satisfied. For the periodic property 5, we have $$\frac{f_2(x+c)}{f_1(x+c)}=1+\sin^2\left(\frac{\pi(x+c)}c\right)=1+\sin^2\left(\frac{\pi x}c+\pi\right)=\frac{f_2(x)}{f_1(x)}.$$ For property 4, it follows $f_1'(x)<0$ and $$f_2'(x)=\frac\pi ce^{-\pi x/c}\left(-\sin^2\left(\frac{\pi x}c\right)+2\sin\left(\frac{\pi x}c\right)\cos\left(\frac{\pi x}c\right)-1\right).$$ Since $\sin^2(t)-2\sin(t)\cos(t)+1\geq0$ for all $t\geq0$, we have $f_2'(x)\leq0$ and thus property 4 is verified.

---

Plot on Desmos