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) $|h^{\prime\prime}(x)| &lt; 1$ for all $x \in (-1,0)\cup(0,1)$

(v) $|h(x)| &gt; \frac{1}{2}$ 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!
Let $h$ satisfy (i)-(iv) and $x_0$ be a point where the maximum of $|h|$ is attained on $[-1,1]\;$. WLOG we can assume that $x_0\ge0$. Then
$$
|h(x_0)|=|h(x_0)-h(1)|=\left|\int_{x_0}^1h'(y)\,dy\right|=
$$
$$
\left|\int_{x_0}^1\int_{x_0}^yh''(z)\,dz\;dy\right|\le
\sup_{(0,1)}|h''|\int_{x_0}^1\int_{x_0}^y\,dz=\frac{(1-x_0)^2}2\sup_{(0,1)}|h''|\le \frac12.
$$
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 $b &gt;0$ 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 $h(0)=1-\frac{1}{8e} &gt; 1/2$. So (iv) and (v) also hold.