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

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

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 $$f(x) \geq 0$$.

• 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. Dec 21, 2019 at 15:25
• 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 $f(x)\ge 0$. Dec 21, 2019 at 16:40
• @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 $x\to-\infty$. Dec 21, 2019 at 16:43
• @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. Dec 21, 2019 at 17:00
• 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. Dec 21, 2019 at 17:06

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

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

or

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

or

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

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

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

or

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

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

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

$$\beta = 0 \tag{19}$$

and

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

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

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

$$OE\Delta$$.

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.