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Is there a $C^{2}$-function $f:\mathbb{R}\to\mathbb{R}$ that is bounded and such that $f'(x)$ is unbounded, but $f''(x)$ is bounded again? For example, $f(x)=\sin(x^2)$ is bounded and has unbounded derivative $f'(x)$, but its second derivative is also unbounded.

edit: Thanks for the great answer. The reason I came up with this question, was the following:

I'd like to find a bounded continuous function $f:\mathbb{R}\to\mathbb{R}$ such that

$\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi t}}e^{-(y-x)^2/2t}f(y)dy -f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-y^2/2}(f(x+\sqrt{t}y)-f(x))dy$

does NOT uniformly converge to $0$ for $t\to 0+$. Any help would be much appreciated.

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Very good question! –  Pete L. Clark May 13 '11 at 3:19
It would be interesting to investigate restrictions leading to the affirmative. –  AD. May 13 '11 at 6:40
@Pete,Steve: +1, Yes. Exactly the one which was in my mind couple of days ago. –  user9413 May 13 '11 at 7:29
Exercise 5.15 in Rudin's Principles of Mathematical Analysis is a quantitative version of this problem. (In a rare moment of candor, Rudin sketches an argument--- basically Giuseppe Negro's argument below, but adapted to his textbook's formulation of Taylor's theorem, and with a slightly better constant. An example is given showing that his constant is in a certain sense the best possible.) –  leslie townes Oct 23 '11 at 22:10
Someone suggested an edit that replaced $\sin x^2$ with $\sin \frac1{x^2}$, saying that the former doesn't have unbounded derivative. To them, two things: (1) given that the domain is all of $\mathbb R$, yes it does, and (2) you should leave a comment instead of making unilateral changes that potentially change the meaning of the question. –  Rahul Jul 27 '12 at 13:34

2 Answers 2

up vote 37 down vote accepted

No, this can't occur. Suppose $f'(x)$ were unbounded but $|f''(x)| < M$ for some $M$. Then for any $N$ you could find some $x_n$ with $|f'(x_n)| > N$. By the mean value theorem for any $y \neq x_n$ one has $$|f'(y) - f'(x_n)| < M|y - x_n|$$ So if $y$ is such that $|y - x_n| < {N \over 2M}$ then $$|f'(y)- f'(x_n)| < M {N \over 2M} = {N \over 2}$$ Since $|f'(x_n)| > N$, this would mean $|f'(y)| > {N \over 2}$, and furthermore by continuity of $f'$, one necessarily has that $f'(y)$ has the same sign as $f'(x_n)$. So integrating one has $$\left|f\left(x_n + {N \over 2M}\right) - f(x_n)\right| = \left|\int_{x_n}^{x_n + {N \over 2M}} f'(y)\,dy\,\right|$$ $$> \left|\int_{x_n}^{x_n + {N \over 2M}} {N \over 2}\,dy\,\right|= {N^2 \over 4M}$$ By the triangle inequality, $|f(x_n + {N \over 2M}) - f(x_n)| \leq |f(x_n + {N \over 2M})| +|f(x_n)|$. So by the above equation, at least one of $|f(x_n + {N \over 2M})|$ and $|f(x_n)|$ is greater than ${N^2 \over 8M}$. You can do this for any $N$, so $f(x)$ must be unbounded.

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+1. Good thinking, Zarrax. –  user9413 May 13 '11 at 7:30
On the other hand, without needing $f$ to be bounded ... –  Glen Wheeler May 13 '11 at 8:13
@Glen: You mean $f$ and $f'$ unbounded and $f''$ bounded? This can happen, just look at $x \mapsto x^2$ from the real line to itself. –  Gunnar Þór Magnússon May 14 '11 at 13:12
@Gunnar indeed, that being my point. (This is a weak "sharpness" observation...) –  Glen Wheeler May 14 '11 at 18:17
There is a minor computational error in the last integral: we have $$\left\lvert \int_{x_n}^{x_n+\frac{N}{2M}} \frac{N}{2}\, dx \right\rvert = \frac{N^2}{4M}$$ you wrote $\frac{N^2}{2M}$. From this we can conclude that at least one of $\lvert f(x_n+\frac{N}{2M})\rvert, \lvert f(x_n) \rvert$ is greater than $\frac{N^2}{8M}$, you wrote $\frac{N^2}{4M}$. It is a totally minor mistake. –  Giuseppe Negro Jun 10 '11 at 11:12

I would like to propose an alternative proof. Let $f \in C^2(\mathbb{R})$ and suppose that $f$ and $f''$ are bounded:

$$\lvert f(x) \rvert \le C_0, \quad \lvert f''(x) \rvert \le C_2.$$

Then $f'$ is bounded. To see this fix $x\in \mathbb{R}$ and write a first-order Taylor expansion of $f$ around $x$:

$$f(x+h)-f(x)= f'(x)h+\int_{x}^{x+h} f''(t)(x+h-t)\, dt,$$

obtaining the estimate

$$\lvert f'(x)\rvert \le \frac{2C_0}{h}+C_2h,\qquad \forall h>0.$$

The right hand side is attains its minimum value for $h=\sqrt{2\frac{C_0}{C_2}}$, so

$$\lvert f'(x)\rvert \le 2\sqrt{2} \sqrt{C_0C_2}.$$

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