Let $X$ be a locally compact Hausdorff space and $K \in L^{2}(X \times X)$, and $\mu$ a positive borel measure.
Define an operator of $L^{2}(X)$, $T(f(x)) = \int_{X} K(x,y) f(y) \mathrm{d}y$.
Decompose $K(x,y) = \sum_{i=1}^{\infty} \psi_{i}(x) \overline{\phi_{i}(y)}$, where $\psi_{i}(x) = \int_{X} K(x,y) \phi_{i}(y) \mathrm{d}y$, where $\{ \phi_{i}\}$ is an orthonormal basis of $L^{2}(X)$.
Show that $\sum_{i=1}^{\infty} |\psi_{i}(x)|^{2} = \int_{X} \int_{X} |K(x,y)|^{2} \mathrm{d}x\mathrm{d}y$
I am having trouble showing that equality. We have $$ \int_{X} \int_{X} |K(x,y)|^{2} \mathrm{d}x\mathrm{d}y = \int_{X} \int_{X} K(x,y) \overline{K(x,y)} \mathrm{d}x\mathrm{d}y \\ = \int_{X} \int_{X} \left( \sum_{i=1}^{\infty} \psi_{i}(x) \overline{\phi_{i}(y)} \right) \left( \overline{\sum_{i=1}^{\infty} \psi_{i}(x) \overline{\phi_{i}(y)}} \right) \mathrm{d}x\mathrm{d}y \\ = \int_{X} \int_{X} \left( \sum_{i=1}^{\infty} \psi_{i}(x) \overline{\phi_{i}(y)} \right) \left( \sum_{i=1}^{\infty} \overline{\psi_{i}(x)} \phi_{i}(y) \right) \mathrm{d}x\mathrm{d}y \\ = \int_{X} \int_{X} \sum_{i,j=1}^{\infty} \psi_{i}(x) \overline{\phi_{i}(y)}\overline{\psi_{j}(x)} \phi_{j}(y) \mathrm{d}x\mathrm{d}y \\ = \int_{X} \sum_{i,j=1}^{\infty} \psi_{i}(x) \overline{\psi_{j}(x)} \int_{X} \left( \overline{\phi_{i}(y)} \phi_{j}(y) \right) \mathrm{d}x\mathrm{d}y \\ = \int_{X} \sum_{i,j=1}^{\infty} \psi_{i}(x) \overline{\psi_{j}(x)} \mathbb{1}_{i=j} \mathrm{d}x \\ = \int_{X} \sum_{i=1}^{\infty} |\psi_{i}(x)|^{2} \mathrm{d}x $$ which is different from the desired equality, is this computation wrong or is the equality and in either case why?