If $f'$ is periodic, $f(0)>0$ and $f' < 1$ show that $f(c)=c$ for some c Let $f \in C^1[0,+\infty)$ where for every $x$ from $[0,+\infty)$, $f'(x+T)=f'(x)$, $T>0$. If $f(0)>0$ and $f'(x)<1$ for every $x$ from $[0,+\infty)$. Prove that there exits $c$ from $[0,+\infty)$ such that $f(x)=x$.
My strategy here is proof by contradiction. I assume that for every $x$, $g(x)=f(x)-x>0$. Hence first derivative is $g'(x)=f'(x)-1<0$. Since $g(0) > 0$ and f is strictly decreasing it follows that there exists $\lim_{x\to \infty}g(x)$. And now i am stuck i don't know how to derive contradiction form this. 
If it were case that $f$ is periodic then i could show that $\lim_{x\to \infty}g(x)$ doesn't exists, hence contradiction. 
Or if $f$ was twice differentiable then i could take second derivative of $g$, which would be periodic because $g'$ is periodic hence it would be the case that exists $\lim_{x\to \infty}g'(x)$ which would be a contradiction since  $g'(x)$ is periodic function.
 A: Note that $f((k+1)T)-f(kT)=\int_{kT}^{(k+1)T}f'(x)\,\mathrm dx=\int_{0}^{T}f'(x)\,\mathrm dx=:c$ is a  constant $<T$, hence for $k$ large enough  $f(kT)=f(0)+kc<kT$. The rest follows by the IVT.
(Or for $g$ instead of $f$: You have a decreasing sequence $g(kT)$ that falls below $0$)
A: Let $\varepsilon >0$ be such that $$\sup_{t\in [0,T]} f'(t) =\sup_{t\in [0,\infty)} f'(t) =1-2\varepsilon .$$
By mean value theorem we have $$\frac{f\left(\frac{f(0)}{\varepsilon}\right) -f(0)}{\frac{f(0)}{\varepsilon}} =f'(\vartheta)\leq 1-2\varepsilon$$ hence $$f\left(\frac{f(0)}{\varepsilon}\right) <\frac{f(0)}{\varepsilon}. $$
So if we define a function $g(x) =f(x) -x$ then we have $$g(0)\cdot g\left(\frac{f(0)}{\varepsilon}\right)<0,$$ hence there exists $c$ such that $g(c) =0.$
A: As $f$ is continuous and periodic, it is bounded, say $f(x)<M$. Hence $f(M)<M$. But $f(0)>0$, and by virtue of the Intermediate Value Theorem, there exists a $c\in(0,M)$, such that
$$
f(c)=c.
$$
Note. The fact that $f$ is differentiable was not used, neither that $f'(x)<0$.
