Find a function that satisfies the following five conditions. My task is to find a function $h:[-1,1] \to \mathbb{R}$ so that
(i) $h(-1) = h(1) = 0$
(ii) $h$ is continuously differentiable on $[-1,1]$
(iii) $h$ is twice differentiable on $(-1,0) \cup (0,1)$
(iv) $|h^{\prime\prime}(x)| < 1$ for all $x \in (-1,0)\cup(0,1)$
(v) $|h(x)| > \frac{1}{2}$ for some $x \in [-1,1]$
The source I have says to use the function 
$h(x) = \frac{3}{4}\left(1-x^{4/3}\right)$ 
which fails to satisfy condition (iv) so it is incorrect. I'm starting to doubt the validity of the problem statement because of this. So my question is does such a function exist? If not, why? Thanks!

Such a function does exist. Try $h(x) = e^{-x^2b}-\frac{b}{e}$. This function trivially satisfies (i)-(iii). Now you just need to find $b$ such that (iv) and (v) also hold. Assuming $b >0$ we get :
$|h^{''}(x)|= |2be^{-x^2b}(2bx^2-1)| \leq 2b |2b-1|$ for $x\in[-1,1]$.
So choosing a very small $b$ like $b=1/8$ will make this smaller than $1/2$. For this $b$ we also get $h(0)=1-\frac{1}{8e} > 1/2$. So (iv) and (v) also hold.
