The property of Hölder continuous density function

Question

I am reading the article but I don't understand some lines in page 10.

First, assumptions on the density $h$ are needed. Of course, since $h$ is the density of a symmetric probability distribution on $[0,1]$, it satisfies $\int h(x) d x=1$ and $\int x h(x) d x=1 / 2$. For the proofs, we will also need the following condition.

Assumption 1. The function $h$ is of class $\mathcal{C}^\beta$ on $[0,1]$, for some $\beta>3$ : the function $h$ is $[\beta]$ times differentiable (where $[\beta]$ is the largest integer smaller than $\beta$ ) and the derivative of order $[\beta]$ is $\beta-[\beta]$ Hölder continuous.

Moreover, we assume that there exists a positive integer $v_0 \geq 2$ such that for all $k \in\left\{0, \ldots, v_0\right\}$, $h^{(k)}(0)=0$

Under Assumption $1, h$ can take positive values only on $(0,1)$, and the function $g$ introduced previously is supported on $\mathbb{R}_{-}$.

I don't understand the reason why under assumption 1, $h$ must take positive values only on (0, 1). Thank you for reading my question and your kind support

Answer 1

$h$ is defined on $[0,1]$, so saying that it "can take positive values only on $(0,1)$" is the same as saying that $h(0) = h(1) = 0$.

The relevant part of the assumption is this one :

Moreover, we assume that there exists a positive integer $v_0\ge 2$ such that $h^{(k)}(0)=0$ for all $k\in\{0,\ldots,v_0\}$

This tells you in particular that $h(0)=0$. Then, because $h$ is symmetric about its mean, it follows that $h(1) = h(1/2 + 1/2) = h(0) = 0 $, exactly as the authors claim.

(Remark : this had nothing to do with Hölder continuity of $h$ !)