Second derivative of bump function? My question is?
Does there exist a function $f\in C^{2}\left(\mathbb{R}\right)$ such that 
$$\begin{cases}
f\left(x\right) & =1\quad\textrm{when }\left|x\right|<1,\\
f\left(x\right) & =0\quad\textrm{when }\left|x\right|\geq2,\\
f\left(x\right) & \in\left[0,1\right],\forall x\in\mathbb{R},\\
\left|f'\left(x\right)\right| & \leq2,\forall x\in\mathbb{R},\\
\left|f"\left(x\right)\right| & \leq2,,\forall x\in\mathbb{R}.
\end{cases}$$
Thanks.

Suppose such an $f$ exists. Note that the hypotheses imply $f'(1)=0=f'(2).$ Consider the line connecting $(1,1)$ and $(2,0).$ This line has slope $-1$. Because $f'(1) = 0,$ We have $f(1+h)$ above this line for small $h>0.$ It follows that for any such $h,$the line connecting $(1+h,f(1+h))$ to $(2,0)$ has slope less than $-1.$ That slope, by the mean value theorem, equals $f'(c)$ for some $c\in (1,2).$ Conclusion: $f'(c) < -1$ for some $c\in (1,2).$
Suppose $c\in (1,3/2].$ The mean value theorem then gives
$$|f''(d)| = |\frac{f'(c)-f'(1)}{c-1}| > \frac{1}{c-1}$$
for some $d \in (1,c).$ Because $0<c-1\le 1/2,$ we see $|f''(d)| >2,$ contradiction. A similar argument works if $c\in [3/2,2).$ Thus there can be no such $f.$
