Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Exercise Let $f\colon \mathbb{R} \to \mathbb{R}$ be $C^2$ and nonnegative. Prove that

$$\big( f'(x)\big)^2 \le 2f(x) \lVert f''\rVert_{\infty}.$$

I have found this innocent-looking little exercise... but I must admit I'm stuck on it. I have tried various roads, the most promising of them being integration by parts: (Notation: $\Delta_hf(x)=(f(x+h)-f(x))/h$)

$$\frac{1}{h}\int_x^{x+h} (f'(t))^2\, dt = f(x+h)\Delta_hf'(x)+f'(x)\Delta_hf(x)-\frac{1}{h}\int_{x}^{x+h}f(t)f''(t)\, dt;$$

I had hoped to bound this identity from above then have $h \to 0^+$. But I got nowhere.

Other approaches used Taylor expansions. All I got with those was a weaker estimate (that I reported here).

Can somebody give me a hint? I can't stop thinking at this exercise but I've got some work to do! :-)

share|cite|improve this question
Maybe obvious, but did you try $(f^2)''$? – Ilya Jun 10 '11 at 12:04
@girdav: I like the idea but I can't get it to work. In fact, suppose $x_0$ is a point of maximum for $g$. Then $g'(x_0)=2f'(x_0)(f''(x_0)-\lvert f''\rVert_{\infty})=0$, which means that either $f'(x_0)=0$ (nice case: from this we infer $g(x_0)\le 0$) or $f''(x)=\lVert f'' \rVert_{\infty}$ in a whole neighborhood of $x_0$ (bad case). How would you handle the bad case? – Giuseppe Negro Jun 10 '11 at 19:12
up vote 2 down vote accepted

The basic idea here is that if the inequality did not hold at some $x_0$, then in an interval centered at $x_0$, $||f''||_{\infty}$ is too small to prevent the graph from descending beyond the $x$-axis in the direction in which it is decreasing.

One can make this rigorous using Taylor expansions... suppose $x_0$ is such that $(f'(x_0))^2 > 2 f(x_0)||f''||_{\infty}$. By a second order Taylor expansion you have $$f(x_0 + h) \leq f(x_0) + f'(x_0)h + {1 \over 2}||f''||_{\infty}h^2$$ If you minimize the right-hand side with respect to $h$ and plug in the condition that $(f'(x_0))^2 > 2 f(x_0)||f''||_{\infty}$ you end out with $f(x_0 + h) < 0$, which contradicts that your function is supposed to be nonnegative.

share|cite|improve this answer
Great! This one works perfectly and it is clear from an intuitive standpoint too. We can also strengthen the result a little: we have $$f'(x)^2\le 2f(x) \sup_{x \in \mathbb{R}}f''(x)$$ which is stricter than $$f'(x)^2 \le 2f(x) \lVert f''\rVert_{\infty}.$$ – Giuseppe Negro Jun 10 '11 at 19:31

I write here the polished version of the answer. This is essentially Zarrax's idea.

Let $S=\sup_{x \in \mathbb{R}} f''(x)$. Since $f\ge 0$, $S\le 0$ implies that $f$ is constant, by concavity. This case is trivial so let us assume $S > 0$. For every $x, h \in \mathbb{R}$ we have

$$0 \le f(x+h)=f(x)+f'(x)h+\int_x^{x+h}f''(t)(x+h-t)\, dt \le f(x)+f'(x)h+\frac{1}{2}Sh^2.$$

For fixed $x$, the last term is quadratic in $h$ with positive leading term. So it is positive iff its discriminant is negative, that is

$$f'(x)^2 \le 2 S f(x),$$

which is the desired inequality.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.