Why is the derivative of the bounded variation nonnegative definite? The following is from a proof I am reading.
Let $C=((c_{ij}))$ be a continuous, symmetric, $d\times d$ matrix-valued function, defined on $[0,\infty)$, satisfying $C(0)=0$ and
$$\sum(c_{ij}(t)-c_{ij}(s))\xi_i\xi_j\geq0\text{, }\xi\in \mathbb R^d\text{, }t>s\geq0.$$
Set
$$\gamma(t)=\sum_{i=1}^d c_{ii}(t)$$
by take $\xi_i=1$ and $\xi_j=0$ otherwise for each $i$, I know $\gamma(t)$ is increasing continuous function. By take $xi_i=1$, $\xi_j=\pm 1$, and $\xi_l=0$ otherwise, 
$$|c_{ij}(t)-c_{ij}(s)|\leq\gamma(t)-\gamma(s)$$
hence $c_{ij}$ is of bounded variation and $c_{ij}$ can be written as
$$c_{ij}(t)=\int_0^t d_{ij}(s)d\gamma(s)$$
My question is why it can conclude $D(s)=((d_{ij}(s)))$ is nonnegative definite? Thanks very much!
 A: Take $a<b$ two real numbers. 
$\displaystyle \sum(c_{ij}(b)-c_{ij}(a))\xi_i\xi_j\geq0 \Longleftrightarrow   \sum \int_a^b d_{ij}(s)\xi_i\xi_j\mathrm d\gamma (s) \geq 0 \Longleftrightarrow  \int_a^b \sum d_{ij}(s)\xi_i\xi_j\mathrm d\gamma (s) \geq 0$
Since $a$ and $b$ are arbitrary we can deduce that $\sum d_{ij}(s)\xi_i\xi_j$ is positive $\mathrm d \gamma(s)$-almost everywhere. This is a classical result from integration theory. Since $\xi$ was also arbitrary $(d_{ij}(s))$ is nonnegative semidefinite almost everywhere (for $\mathrm d \gamma (s))$.
Note that if $\gamma$ is constant on $[a,b]$ then $\mathrm d \gamma (s)$ is identically zero on $[a,b]$ and $(d_{ij}(s))$ can be wathever you want(even a negative definite matrix).
However $(d_{ij})$ does not need to be nonnegative definite. One can have (for fixed $a$ and $b$) $C(b)-C(a)$ to be non zero, nonnegative semidefinite but not nonnegative definite and with $\gamma(b)-\gamma(a)>0$. Plus you can find $C$ satifsfying all that plus the fact that there exists a non-zero $\xi$ such that $(C(t)\xi,\xi)=0$ for all $t\in[a;b]$. Then $\mathrm d \gamma$ is not zero on $[a;b]$ and $\displaystyle \sum(c_{ij}(b)-c_{ij}(a))\xi_i\xi_j=0$ so $(D(t))$ is only nonnegative semidefinite $\mathrm d \gamma$ almost everywhere and not definite.
