Why is it true that if 
$f(x):\mathbb{R}\longrightarrow\mathbb{R}$ is a positive, even function, decreasing for x>0, then it can be written as a convex linear combination of $\frac{1}{2h}\chi\_{[-h,h]}(x)$? 
Thank you, I'm struggling with this a lot!



Even function and linear combination
Suppose that $f\colon\mathbb{R}\longrightarrow\mathbb{R}$ is a positive even function, which is decreasing and left-continuous on $[0,\infty)$, and such that $f(x)\to0$ as $x\to\infty$. Then for real $t\ge0$
\begin{multline}
 f(t)=\int\_{[t,\infty)}-df(u)
=\int\_{[0,\infty)}-df(u)1\_{t\le u} \\ 
=\int\_{(0,\infty)}-df(u)1\_{t\le u}
=\int\_{(0,\infty)}-df(u)1\_{|t|\le u} \\ 
=\int\_{(0,\infty)}-df(u)\chi\_{[-u,u]}(t)
=\int\_{(0,\infty)}\mu(du)\frac{\chi\_{[-u,u]}(t)}{2u}, \tag{0}
\end{multline} 
where the measure $\mu$ is defined by the formula $\mu(du)=-2u\,df(u)$. 
In display (0), the first equality holds because $f$ is left-continuous on $[0,\infty)$ and $f(x)\to0$ as $x\to\infty$; the second equality holds because $t\ge0$; the third equality holds because $f$ is left-continuous at $0$ and, being even, is also right-continuous at $0$ and hence continuous at $0$, so that $\int\_{\{0\}}df(u)=0$; and the fourth equality holds because $t\ge0$; the remaining equalities in (0) are trivial. 


So, 
\begin{equation\*}
 f(t)=\int\_{(0,\infty)}\mu(du)\frac{\chi\_{[-u,u]}(t)}{2u}
\end{equation\*}
for $t\ge0$. Since $f$ is even, the latter identity holds for all real $t$. 
So, $f$ is indeed a positive mixture of functions $\frac{\chi\_{[-u,u]}}{2u}$ with $u>0$. 


Also,
\begin{multline\*}
 \int\_0^\infty \mu(du)
=\int\_0^\infty -2u\,df(u)
=\int\_0^\infty -2\,df(u)\int\_{(0,u]} dt \\ 
=\int\_0^\infty dt\int\_{[t,\infty)}-2\,df(u)
=\int\_0^\infty dt\,2 f(t)=\int\_{-\infty}^\infty dt\,f(t);
\end{multline\*}
the third equality here is an instance of the Fubini--Tonelli theorem. 


So, the positive mixture is actually a convex mixture (that is, $\int\_0^\infty \mu(du)=1$) if and only if 
\begin{equation}
 \int\_{-\infty}^\infty du\,f(u)=1. \tag{1}
\end{equation}




---


It is clear that the condition $f(x)\to0$ as $x\to\infty$ is necessary for (1), given that $f$ is positive and decreasing on $[0,\infty)$. 


Also, the example of the function $f=\frac{\chi\_{(-1,1)}}2$, which cannot be represented as a positive mixture of functions $\frac{\chi\_{[-u,u]}}{2u}$ with $u>0$, shows that the condition that $f$ be left-continuous on $[0,\infty)$ cannot be dropped. 


