Let $f$ be $C^1$ function on $\mathbb{R}$. Suppose that $$f'(x)\leq 1-f(x)^2$$ Prove that $|f(x)|\leq 1 \: \forall x$.
I suppose that such a function can be bounded by a solution to the ODE: $y'=1-y^2$ with the same initial value, but I don't know by which theorem, if at all.
Trying to solve this ODE gives: $$\frac{dy}{1-y^2}=dx \Rightarrow \int\frac{dy}{1-y^2}=\int dx=x+c$$ but now another problem arises: $\frac{1}{y^2-1}=\frac{1}{2}\left ( \frac{1}{y-1}-\frac{1}{y+1} \right )$ and so taking anti-derviative depends whether $|y|\leq 1$ or not, which makes me even more confused.
Can someone please make me some sense of all this mess?
It is sufficient to prove that $$|f(0)|=:c>1\tag{1}$$ is impossible.
If $(1)$ holds with $f(0)=c$ define $g(x):=f(-x)$, and if $(1)$ holds with$f(0)=-c$ define $g(x):=-f(x)$. In both cases we obtain $$g(0)=c>1,\qquad g'(x)\geq g^2(x)-1\quad(x\geq0)\ .$$ It is then obvious that $g$ increases on ${\mathbb R}_{\geq0}$; hence $g(x)>1$ for all $x\geq0$. Now $$g'\geq g^2-1=(g-1)^2+2(g-1)> (g-1)^2$$ implies $${g'(t)\over \bigl(g(t)-1\bigr)^2}>1\qquad(t\geq0)\ .$$ We integrate this from $t=0$ to $t=x>0$ and obtain $$-{1\over g(t)-1}\biggr|_0^x> x\ ,$$ and this expands to $$g(x)>1+{1\over C-x}\quad (0<x<C)\ ,$$ where we have put ${1\over c-1}=:C$. This shows that $g$, hence $f$, cannot live over all of ${\mathbb R}$.