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!