bounding a $C^1$ function that upholds $f'(t)\leq 1-f(t)^2$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function which is continuously differentiable.

Suppose that $\forall t\in\mathbb{R}$ it holds $$f'(t)\leq 1-f(t)^2$$
Prove that $\forall t\in\mathbb{R}:|f(t)|\leq 1$.

I know about some comparison theorems, i.e. if we have a solution (in a compact domain) of the Riccati equation $\dot{x}(t)=1-x(t)^2$, with the same initial value as of $f$, then in this domain we get that $f\leq x$. My problem is that it only gives an upper bound and not a lower bound.

Answer 1

The key point to consider is that the domain of the function is demanded to be the whole of $\Bbb R$. The function is not allowed to have poles or other kinds of discontinuities, it has to stay bounded over all finite intervals.

Claim: If $f$ leaves the interval $[-1,1]$, $|f(t_*)|>1$ for some $t_*$, then $f$ has a pole, that is, it has to diverge towards $\pm \infty$ in finite time.

If $f(t_*)<-1$, the function has to be monotonically falling around $t_*$ and keep being falling for $t>t_*$. Then conclude that for $t>t_*$ $$ f'(t)\le -cf(t)^2~~\text{ where }~~c=\frac{f(t_*)^2-1}{f(t_*)^2} $$ so that $$ f(t)\le\frac{f(t_*)}{1+cf(t_*)(t-t_*)} $$ goes to minus infinity in finite time. This contradicts the domain being the whole of $\Bbb R$, so that $f$ has to be bounded on finite intervals.

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Or put another way, by assumption $f(t)$ is finite and by the general considerations above falling for $t\ge t_*$, so that $f(t)<f(t_*)<-1$. By the mean value theorem one can now conclude $$ -\frac1{f(t_*)}\ge-\frac1{f(t_*)}+\frac1{f(t)}=-\frac{f'(s)}{f(s)^2}(t-t_*)\ge\left(1-\frac1{f(s)^2}\right)(t-t_*)\ge\left(1-\frac1{f(t_*)^2}\right)(t-t_*) $$ which is quite impossible as the left side is a positive constant while the right is an unbounded increasing function.

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The same argument goes for $f(t_*)>1$ looking in direction $-\infty$ and growth towards $+\infty$, since $\tilde f(t)=-f(-t)$ satisfies the same assumptions.

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Added 7/2019: Use the structure of the equality solution

Assume again $f(t_*)<-1$ and consider $$g(t)=\dfrac{f(t)+1}{f(t)-1}=1+\dfrac2{f(t)-1}.$$ At $t=t_*$ we get $g(t_*)\in(0,1)$. It is not possible for $g$ to have the value $1$.

Now consider the derivative and insert the differential inequality $$ g'(t)=-\frac{2f'(t)}{(f(t)-1)^2}\ge -2\frac{1-f(t)^2}{(f(t)-1)^2}=2g(t). $$ By the general Grönwall theory we get for $t>t_*$ $$ g(t)\ge e^{2(t-t_*)}g(t_*). $$ This grows unbounded and thus has to cross the value $1$. At that point $f$ must have a singularity.