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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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    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.2012-07-27

2 Answers 2

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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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    nice proof have you think about using weak derivative or almost pointwise derivatives?2012-07-16
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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 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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    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.134.2017-08-23