Show that there is no differentiable function $f :\Bbb R \to\Bbb R$ such that $f(0) = 1$ and $f'(x) ≥ (f(x))^2\space\forall x\in\Bbb R$

Question

Show that there is no differentiable function $f :\Bbb R \to\Bbb R$ such that $f(0) = 1$ and $f'(x) ≥ (f(x))^2\space\forall x\in\Bbb R$.

My attempt:

Suppose there is a differentiable function such that the above statement holds true.

Then, as the function is differentiable so it will satisfy the mean value theorem. $$f(x)-f(0)=f'(c)(x-0),x>0,c \in (0,x)$$

$$f(x)-1=f'(c)x$$ $$f(x)-1 \ge (f(c))^2 x$$ $$f(x)-(f(c))^2 x -1 \ge 0$$

If we let $h(x) = f(x)-(f(c))^2.x-1$ then $h'(x) \ge 0$ then $f'(x)-(f(c))^2 \ge 0$ . How do I proceed after this and obtain a contradiction.

Answer 1

We may assume $f$ never vanishes for $x>0$. Because if there exists a point $x_0$ such that $f(x_0)=0$, then by the mean-value theorem, one has a point $c\in (0,x_0)$ such that \begin{align*} f(x_0)-f(0) = f’(c)x_0\Rightarrow -1 \ge (f(c))^2 x_0 \ge 0 \end{align*} which is not possible. So assume $f(x)\neq 0$ for all $x>0$. Now, if $f’(x) \ge (f(x))^2$ is true, then one has \begin{align*} \frac{f’(x)}{f(x))^2} \ge 1 & \Rightarrow -\frac{1}{f(x) }\ge x + c \end{align*} for some constant $c$. But for large enough $x>0$, we have $f(x) < 0$.

Since $f(0)=1 >0$, then by the intermediate value theorem, we must have $f(x’)=0$ for some $x’>0$. But we already assumed $f$ never vanishes.

Answer 2

Consider the function $g(x) = x+\frac{1}{f(x)}$.

Note $g'(x)= 1-\frac{f'(x)}{f^2(x)} \leq 0$. Hence $g(x)$ is a non-increasing function.

But $f'(x) \geq f^2(x) \geq 0$ implies $f(x)$ is non-decreasing function.

Since, $f(0) = 1$ implies $f(x) \geq 1 \geq 0$ (as $f$ is non-decreasing$).

Hence $g(100) \geq 100 $ but $g(x)$ is a non-increasing function implies $g(100) \leq f(0) = 1$. This a contradiction.