TITLE: Find a function that satisfies the following five conditions.
QUESTION [3 upvotes]: 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!

REPLY [3 votes]: 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
$$
|h(x_0)|=|h(x_0)-h(1)|=\left|\int_{x_0}^1h'(y)\,dy\right|=
$$
$$
\left|\int_{x_0}^1\int_{x_0}^yh''(z)\,dz\;dy\right|\le
\sup_{(0,1)}|h''|\int_{x_0}^1\int_{x_0}^y\,dz=\frac{(1-x_0)^2}2\sup_{(0,1)}|h''|\le \frac12.
$$
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