Proving that $f'(x)\le f(x), \forall x\in \mathbb{R}$

Let $f:\mathbb{R} \to \mathbb{R}$ be a twice differentiable function such that

$0 \le f''(x) \le f(x)$
$f'(x)\ge 0, \forall x\in \mathbb{R}$.

Thus, prove that $f'(x)\le f(x), \forall x\in \mathbb{R}$.

I couldn't make much progress, but I observed that $f$ is increasing and convex while $f'$ is increasing from the hypothesis.

Then I tried to evaluate the derivative of $h:\mathbb{R} \to \mathbb{R}$ with $$ h(x)=f(x)-f'(x)$$

and I got that $$h'(x)=f'(x)-f''(x), \forall x\in \mathbb{R}$$

Now, I would be done if I could prove that $h'(x)\ge 0$ for all $x\in \mathbb{R}$. But I'm not sure yet if this approach is really that straightforward.

I also tried to use that $f$ is convex, but to no avail.

EDIT: Maybe that we should somehow use the fact that a differentiable function $g:\mathbb{R}\to\mathbb{R}$ is convex if and only if $g(x)\ge g(y)+g'(y)(x-y),\ \forall x,y\in \mathbb{R}$

Solution 1:

For a non-negative, increasing, convex function, we have $\lim_{x \to -\infty}(f(x) - f'(x)) \geq 0$ exists. Consider: $$ g(x) = (f(x) - f'(x))e^x $$ Differentiating yields: $$ g'(x) = (f(x) - f''(x))e^x \geq 0 $$ We have that $\lim_{x \to -\infty} g(x) \geq 0$. Since $g'(x) \geq 0$, $g$ is increasing so this implies $g(x) \geq 0$ $\forall x \in \mathbb{R}$. Since $e^x > 0$, we have $f(x) \geq f'(x)$.

EDIT: To prove that $\lim_{x \to -\infty}(f(x) - f'(x)) \geq 0$, we shall show that $\lim_{x \to -\infty} f(x) \geq 0$ and $\lim_{x \to -\infty} f'(x) = 0$.

$\lim_{x \to -\infty} f(x)$ exists because of monotone convergence theorem. Take any $(x_n)_{n \in \mathbb{Z}^+}$ such that $x_n \to -\infty$, and we may assume WLOG that $x_n$ is monotone (as otherwise we can remove the terms which violate monotonicity, and the asymptotic behavior remains unchanged). Since $x_n \geq 0$ $\forall n$, we have that $x_n$ converges. Since $f(x) \geq 0$, the limit must also be non-negative.

Similarly, we can prove that $\lim_{x \to -\infty} f'(x) \geq 0$ as we have $f'(x) \geq 0$ and $f''(x) \geq 0$ so $f'$ is a monotonically increasing function bounded from below. We show further that $\lim_{x \to -\infty} f'(x) = 0$ by contradiction, in which if $\lim_{x \to -\infty} f'(x) = \epsilon > 0$, then for some $M > 0$ we have $f'(x) > \frac{\epsilon}{2}$ for all $x < -M$. This would violate the assumption that $f(x) \geq 0$.

Solution 2:

With

$0 \le f''(x) \le f(x), \; \forall x \in \Bbb R, \tag 1$

and

$f'(x) \ge 0, \; \forall x \in \Bbb R, \tag 2$

we have

$0 \le f'(x) f''(x) \le f(x) f'(x), \; \forall x \in \Bbb R, \tag 3$

$0 \le \dfrac{1}{2} ((f'(x))^2)' \le \dfrac{1}{2} (f^2(x))', \tag 4$

$0 \le ((f'(x))^2)' \le (f^2(x))', \tag 5$

we integrate this 'twixt some arbitrary $L \in \Bbb R$ and $x \in \Bbb R$ to obtain

$(f'(x))^2 - (f'(L))^2 = \displaystyle \int_L^x ((f'(s))^2)' \; ds \le \int_L^x (f^2(s))' \; ds = f^2(x) - f^2(L). \tag 6$

Now in light of (1),

$f(x) \ge 0, \forall x \in \Bbb R, \tag 7$

so if we define

$\alpha = \inf \{ f(x), \; x \in \Bbb R \}, \tag 8$

then

$\alpha \ge 0 \tag 9$

and, by virtue of (2), $f(x)$ is monotonically decreasing with decreasing $x$; together these imply that

$\displaystyle \lim_{x \to -\infty} f(x) = \alpha. \tag{10}$

We next consider $f'(x)$ as $x \to -\infty$; again in accord with (1) we see that $f'(x)$ is monotonically decreasing with decreasing $x$, and by (2) it too is bounded below by $0$. I claim that in fact

$\displaystyle \lim_{x \to -\infty} f'(x) = 0; \tag{11}$

for if not, setting

$\beta = \inf \{f'(x), x \in \Bbb R\} > 0, \tag{12}$

then we may assert that

$f'(x) \ge \beta, \; \forall x \in \Bbb R; \tag{13}$

then picking

$x_0, x_1 \in \Bbb R, \; x_0 < x_1, \tag{14}$

it follows that

$f(x_1) - f(x_0) = \displaystyle \int_{x_0}^{x_1} f'(s) \; ds \ge \int_{x_0}^{x_1} \beta \; ds = \beta(x_1 - x_0), \tag{15}$

whence

$f(x_1) - f(x_0) \ge \beta(x_1 - x_0), \tag{16}$

$f(x_0) - f(x_1) \le -\beta(x_1 - x_0), \tag{16}$