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 $g(x)\ge g(y)+g'(y)(x-y),\ \forall x,y\in \mathbb{R}$

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    $\begingroup$ How does $h' \geq 0$ complete the proof? $\endgroup$ Dec 20, 2019 at 23:53
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    $\begingroup$ Good question. The only example of such functions I can come up with are $f(x)=e^{cx}$ with $ 0\leq c \leq 1$. $\endgroup$ Dec 20, 2019 at 23:55
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    $\begingroup$ @KaviRamaMurthy Yeah, you are right, it doesn't. It would only if I knew that $\lim_{x\to -\infty} h(x)=0$. $\endgroup$
    – Math Guy
    Dec 20, 2019 at 23:57

2 Answers 2


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 $f(x) \geq 0$.

  • $\begingroup$ Thank you, nice proof ! Could you help me to prove that $\lim_{x \to -\infty}(f(x) - f'(x)) \ge 0$? I tried using that $f(x)\ge f(a)+f'(a)(x-a), \forall a,x \in \mathbb{R}$, but I didn't succeed. $\endgroup$
    – Math Guy
    Dec 21, 2019 at 15:25
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    $\begingroup$ I don't think that it works, because from $\lim_{x\to -\infty}f(x)\ge 0$ and $\lim_{x\to -\infty} f'(x)\ge 0$ it doesn't follow that their difference is $\ge 0$. However, if we could show that $\lim_{x\to -\infty} f'(x)=0$ then it would be all right. Yet, I can't see how the fact that for some $M>0$ we have that $f'(x) >\frac{\epsilon}{2}, \forall x<-M$ contradicts that $f(x)\ge 0$. $\endgroup$
    – Math Guy
    Dec 21, 2019 at 16:40
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    $\begingroup$ @MathGuy: I was going to comment the same. However the conclusion $f'(x) \to 0$ as $x\to-\infty $ is true by mean value theorem. Just note that $f(x+1)-f(x) =f'(c) $ and let $x\to-\infty $. $\endgroup$
    – Paramanand Singh
    Dec 21, 2019 at 16:43
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    $\begingroup$ @MathGuy you're right. I initially was thinking of proving $\lim_{x \to -\infty} f'(x) = 0$, but for some reason, I thought proving $\geq 0$ would suffice. I've edited my post. $\endgroup$ Dec 21, 2019 at 17:00
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    $\begingroup$ It is just as per what @ParamanandSingh mentioned. Let $x_0 = -M - 1$, and let $x_{n+1} = x_n - 1$. By MVT, we have $f(x_{n+1}) - f(x_n) = f'(c)(x_{n+1} - x_n) < -\frac{\epsilon}{2}$. This means that if $f(x_0) = y_0$, then $f(x_n) < y_0 - n\frac{\epsilon}{2}$. Choose $n$ sufficiently large and we have $f(x_n) < 0$, the required contradiction. $\endgroup$ Dec 21, 2019 at 17:06


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


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

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

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


$\alpha \ge 0 \tag 9$

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

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


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


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

that is,

$f(x_0) \le f(x_1) - \beta(x_1 - x_0); \tag{17}$

but it is easily seen that this implies that

$f(x_0) \to -\infty \; \text{as} \; x_0 \to -\infty, \tag{18}$

which contradicts (1). Thus

$\beta = 0 \tag{19}$


$f'(x) \to 0 \; \text{as} \; x \to \infty. \tag{20}$

Returning now to (6), we have

$(f'(x))^2 - (f'(L))^2 \le f^2(x) - f^2(L), \tag {21}$

and letting

$L \to -\infty \tag{22}$

we reach

$(f'(x))^2 \le f^2(x) - \alpha^2 \le f^2(x), \tag {23}$

and since both

$f(x), f'(x) \ge 0, \forall x \in \Bbb R, \tag{24}$

we may at last conclude that

$f'(x) \le f(x), \; \forall x \in \Bbb R, \tag{25}$


Finally, note that in light of (10) and (11) we have

$\displaystyle \lim_{x \to -\infty} (f(x) - f'(x))$ $= \lim_{x \to -\infty} f(x) - \lim_{x \to -\infty}f'(x) = \alpha - 0 \ge 0, \tag{26}$

as our OP requested be proved in a comment to Clement Yung's answer.


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