Positive semidefiniteness of the Brownian covariance kernel

Symmetry is immediate from $\min (s,t)=\min (t,s)$. If $n=0$, the displayed quadratic form is the empty sum $0$. Now take $n\ge 1$, let $0=r_0<r_1<\cdots <r_k$ be the increasing list of the distinct positive values among $t_1,\dots ,t_n$, and put $A_{\mathrm{\ell }}={\sum }_{i:t_i\ge r_{\mathrm{\ell }}}{a}_i$. The list is finite and uniquely fixed by the given times; no choice function is used.

For every $i,j$, $\min (t_i,t_j)={\sum }_{\mathrm{\ell }=1}^k(r_{\mathrm{\ell }}-r_{\mathrm{\ell }-1}){\mathbf{1}}_{\{t_i\ge r_{\mathrm{\ell }}\}}{\mathbf{1}}_{\{t_j\ge r_{\mathrm{\ell }}\}}\mathrm{.}$ Indeed, if the smaller of $t_i,t_j$ is $0$ both sides vanish, while if it is $r_m$ the right side telescopes to ${\sum }_{\mathrm{\ell }=1}^m(r_{\mathrm{\ell }}-r_{\mathrm{\ell }-1})=r_m$.

Substituting step 2.1 and rearranging only finite sums gives ${\sum }_{i,j=1}^n{a}_i{a}_j\min (t_i,t_j)={\sum }_{\mathrm{\ell }=1}^k(r_{\mathrm{\ell }}-r_{\mathrm{\ell }-1}){\left({\sum }_{i:t_i\ge r_{\mathrm{\ell }}}{a}_i\right)}^2={\sum }_{\mathrm{\ell }=1}^k(r_{\mathrm{\ell }}-r_{\mathrm{\ell }-1})A_{\mathrm{\ell }}^2\mathrm{.}$ Every weight is positive and every square is nonnegative, so the quadratic form is nonnegative. This includes coincident times, zero times, zero coefficients, and the case $k=0$, where the last sum is empty.