why $\sum_{n=1}^\infty \int_0^c \tau_n(x)dx \int_0^d \overline{\tau_n(x)}dx$ is the $min(c,d)$?

Question

Let $\{ \tau_n (x) \}_{n=1}^\infty$ be closed orthonormal system in $L^2_{pc}[0,1]$ show that for $c,d\in (0,1)$ $$\sum_{n=1}^\infty \int_0^c \tau_n(x)dx \int_0^d \overline{\tau_n(x)}dx = min\{c,d\}$$

Answer 1

Consider the characteristic functions of the intervals $\chi_d : = \chi_{[0,d]}, \chi_C := \chi_{[0,c]}$. Then, we have: $$\sum_{n=1}^\infty \int_0^c \tau_n(x)dx \int_0^d \overline{\tau_n(x)}dx = \sum_{n=1}^\infty \int_0^1 \tau_n(x) \chi_d(x) dx \int_0^1 \overline{\tau_n(x)}\chi_c(x)dx$$ Notice now that for any $f,g$ in $L^2$ we have:

$$ \langle f,g \rangle = \sum_{n = 1}^\infty \langle f, e_n \rangle \langle g , e_n \rangle$$

In particular it follows: $$ \sum_{n=1}^\infty \left[ \int_0^1 \tau_n(x) \chi_d(x) dx \int_0^1 \overline{\tau_n(x)}\chi_c(x)dx \right] = \int_0^1 \chi_c(x) \chi_d(x)dx = \min(c,d) $$ as the left hand side above is the right hand side two lines above, and vice versa