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.
 A: 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.
A: 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.
