The dot product is a valid kernel. Is the integral too?
Solution 1:
First, you are correct about writing the summation as the integral. Look here - Integration with respect to counting measure.
And yes, it is the same as for the dot product kernel because it is the inner product on the space of sequences $l^2$ $$\sum_i^n\sum_j^nc_ic_j\sum_m^\infty y_{i,m}y_{j,m}=\sum_i^nc_i\sum_m^\infty y_{i,m}(\sum_j^n c_jy_{j,m})=\sum_m^\infty (\sum_i^n c_iy_{i,m})(\sum_j^n c_jy_{j,m})=\sum_m^\infty (\sum_i^n c_iy_{i,m})^2\geq0$$ Because finite and infinite sums (or integrals) can be exchanged.