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I know that a bounded function with a bounded second derivative also has a bounded first derivative.

But consider a bounded function $f(x)$ such that $f'(x)$ is also bounded.

Can we claim that $f''(x)$ will also be bounded. Intuition says NO !

But I am unable to find an example. Please help.

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    $\begingroup$ Can you construct a bounded function with unbounded first derivative? If so, then what happens to the integral of that function? $\endgroup$
    – ConMan
    Jan 18 at 5:10
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    $\begingroup$ Look up the Fresnel integral. $\endgroup$
    – copper.hat
    Jan 18 at 5:12
  • $\begingroup$ The function defined by $g(x)=\text{sin}(\text{sin}(x))$ have all derivatives bounded. $\endgroup$ Jan 18 at 5:52
  • $\begingroup$ Aah! It was a typo. Thanks for pointing out. I meant to say $f(x) = \sin(x \sin(x))$. It has an unbounded derivative. $\endgroup$ Jan 18 at 6:02
  • $\begingroup$ I have deleted the previous comment as it has a typing error: Say, we look at $f(x)=\sin(x \sin(x))$. Then it is a bounded function with an unbounded derivative. So its integral will suffice for what I am looking for. But the problem is that I cannot think of a standard mathematical function which can express the integral. To be fair, this example definitely works. But I am looking for a more pretty example. $\endgroup$ Jan 18 at 6:04

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The bounded analytic function $g: \mathbb{R} \rightarrow \mathbb{R}$ defined by $g(x)=\dfrac{\text{sin}(x^3)}{x^2+1}$, for all $x\in\mathbb{R},$ have the first derivative bounded. The derivatives are: $$ \begin{align} g'(x) = & \dfrac{x (3 (x + x^3) \text{cos}(x^3) - 2 \text{sin}(x^3)))}{(1 + x^2)^2} \\ g''(x)= & - \dfrac{6 x (x^4 - 1) \text{cos}(x^3) + (9 x^8 + 18 x^6 + 9 x^4 - 6 x^2 + 2) \text{sin}(x^3) }{(1 + x^2)^3}, \end{align} $$ for all $x \in \mathbb{R}.$

The first derivative is bounded because the largest power multiplying the trigonometric functions in the numerator is 4, which is the same of the degree of the denominator. This means that the limit of these polynomial terms exists, implying that $g'(x)$ is bounded, since sine and cosine are bounded functions.

The second derivative is unbounded because the denominator have degree $6$ while the polynomial term following the sine have degree $8$, two degrees higher. Choosing a sequence of real numbers $\{x^k\}_{k\in\mathbb{N}}$ such that $\sin(x^{k})=1$ and $\cos(x^{k})=0$ diverging to infinity is enough for proving the unboundedness of the second derivative.

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    $\begingroup$ yeah !I got it I gave a wrong example. $\endgroup$
    – user1012971
    Jan 18 at 6:24
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    $\begingroup$ Thanks for pointing it out. $\endgroup$
    – user1012971
    Jan 18 at 6:25
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    $\begingroup$ Just a typo (derined) in the first line though understandable.. $\endgroup$
    – user1012971
    Jan 18 at 6:46
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    $\begingroup$ Brilliant answer. thanks a lot ! $\endgroup$ Jan 18 at 7:47
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    $\begingroup$ Thanks! Your question is good to practice some intuitions we have. For some m, functions of the form $$f(x)= \dfrac{\text{sin}(x^{m})}{x^{2n}+1}$$ seems to have the first n-derivatives bounded while the $n+1$ and higher unbounded. The idea is that the denominator explodes at infinity fast enough to the sine function not to be able to compensate this fast growth. The idea behind the sine function is that it oscillates so that the high order derivatives starts to come into consideration in the Taylor expansion. $\endgroup$ Jan 18 at 18:30

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