Let $A = [a_{ij}(t)]$ be any $n$-by-$n$ square matrix function defined for $t>0$, and $B =[b_{ij}]$, where $$b_{ij} = \int_0^\infty \,a_{ij}(t)\,\text{d}t\,.$$ An $n$-by-$n$ matrix $M$ is said to be conditionally positive definite or CPD if $\langle x, Mx\rangle \geq 0$ for all $x \in \mathbb{C}^{n}$ such that $x_{1}+x_{2}+.......+x_{n}=0$. Then prove that $B$ is CPD, if $A(t)$ is CPD for every $t\in (0,\infty)$."
Any help! I know I need to use inner product definition here, but the integral sign is bothering me. Any suggestion or tip will be really helpful! Thanks in advance.
Suppose that $A(t)=\begin{bmatrix}a_{i,j}(t)\end{bmatrix}_{i,j\in\{1,2,\ldots,n\}}$ is a CPD (Chicago Police Department?) matrix for each $t\in\mathbb{R}_{>0}$. Let $x:=(x_1,x_2,\ldots,x_n)\in\mathbb{C}^n$ be such that $x_1+x_2+\ldots+x_n=0$. Then $$\big\langle x,A(t)\,x\big\rangle\geq 0\,.\tag{*}$$
Observe that $$ \langle x,B\,x\rangle =\Biggl\langle x,\left(\int_0^\infty\,A(t)\,\text{d}t\right)\,x\Biggr\rangle\,,$$ where the integral of a matrix function is done entrywise. By linearity of the inner product and of the integral operator, we see that $$ \langle x,B\,x\rangle =\int_0^\infty\,\big\langle x,A(t)\,x\big\rangle\,\text{d}t\,.$$ From (*), we conclude that $\langle x,B\,x\rangle \geq 0$, whence $B$ is also a Chicaco police officer.