# Is there a bounded function $f$ with $f'$ unbounded and $f''$ bounded?

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.

• Very good question! Commented May 13, 2011 at 3:19
• It would be interesting to investigate restrictions leading to the affirmative. Commented May 13, 2011 at 6:40
• @Pete,Steve: +1, Yes. Exactly the one which was in my mind couple of days ago.
– user9413
Commented May 13, 2011 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.) Commented Oct 23, 2011 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.
– user856
Commented Jul 27, 2012 at 13:34

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.

• On the other hand, without needing $f$ to be bounded ... Commented May 13, 2011 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. Commented May 14, 2011 at 13:12
• @Gunnar indeed, that being my point. (This is a weak "sharpness" observation...) Commented May 14, 2011 at 18:17
• nice proof have you think about using weak derivative or almost pointwise derivatives? Commented Jul 16, 2012 at 18:53

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 $$\lvert f'(x)\rvert \le 2\sqrt{C_0 C_2},$$ so in particular $$f'$$ is bounded.

Proof. We fix $$x\in \mathbb{R}$$ and write a first-order Taylor expansion of $$f$$ around $$x$$; $$f(x+h)-f(x)= f'(x)h+\int_{x}^{x+h} f''(t)(x+h-t)\, dt,$$

so, rearranging terms, we obtain the estimate $$\tag{1}\lvert f'(x)\rvert \le \frac{2C_0}{\lvert h\rvert}+C_2\lvert h\rvert ,\qquad \forall h\ne 0.$$ The right hand side attains its minimum value for $$h=2\sqrt{\frac{C_0}{C_2}}$$, so
$$\lvert f'(x)\rvert \le 2 \sqrt{C_0C_2}.$$ This concludes the proof.

Remark 1. Martin R points out in the comments that the bound can be improved to $$\lvert f'(x)\rvert \le \sqrt 2 \sqrt{C_0C_2}.$$

Remark 2. Here we used that $$f$$ is smooth on the whole real line. If $$f$$ is smooth only on a bounded interval $$I$$, then I think that the bound can be worse. Indeed, in the proof it could be that the choice of $$\lvert h\rvert$$ that minimizes the right-hand side of (1) is not available, because $$x+h$$ falls outside of $$I$$. But I don't have an explicit example.

• The final inequality, with $2\sqrt 2$ replaced by a non-explicit constant, can be obtained as an application of the closed graph theorem. See the book "Functional Analysis", Eidelman - Milman - Tsolomitis, Theorem 9.3.4 pag.134. Commented Aug 23, 2017 at 10:21
• You actually have $\lvert f'(x)\rvert \le \dfrac{2C_0}{h}+\dfrac{C_2h}{2}$. That gives $\lvert f'(x)\rvert \le 2\sqrt{C_0C_2}$ as in math.stackexchange.com/q/257020/42969. – A discrete analogue of this question is currently discussed at math.stackexchange.com/q/3397548/42969. Commented Oct 18, 2019 at 11:36
• @MartinR: Oh, you are right, that improves the constant by a factor of $\sqrt 2$. Is $2\sqrt{C_0C_2}$ the best possible constant? I guess so. Commented Oct 18, 2019 at 12:43
• According to Landau–Kolmogorov inequality, $C(2, 1, \Bbb R) = 2$ is the sharp bound for this case. Commented Oct 18, 2019 at 13:08
• The sharp bound is $|f'(x)| \le \sqrt 2 \sqrt{C_0 C_2}$, compare math.stackexchange.com/q/3622187/42969. Commented Apr 13, 2020 at 11:49