# Exercise: manufacture a bound on $f'$ from $f$ and $f''$

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! :-)

-
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

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.

-
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.

-