$\sum_{n=1}^\infty (AB e_n) = \sum_{n=1}^\infty \sum_{m=1}^\infty (Be_n, e_m) (Ae_m, e_n)$

Question

Let $H$ be a Hilbert space, $A$ and $B$ bounded linear operators on $H$, and $\{e_n\}$ an orthonormal basis for $H$. I am following a proof that claims it is obvious that $$\sum_{n=1}^\infty (AB e_n, e_n) = \sum_{n=1}^\infty \sum_{m=1}^\infty (Be_n, e_m) (Ae_m, e_n).$$ However I do not see why this is necessarily true. How can one prove this?

Answer 1

It basically just uses that any vector $v$ in a seperable Hilbert space $H$ with basis $\{e_1,e_2,...,\}$ can be expressed as $v = \sum\limits_{n=1}^\infty (v,e_n) e_n$.

So one has $B e_n = \sum\limits_{m=1}^\infty (B e_n,e_m)e_m$ and since $A$ is a linear operator, i.e. $A(\lambda v) = \lambda A(v)$ for any $v\in H$ and $\lambda\in\mathbb{R}$, this means that \begin{align*} (AB)e_n &= A(Be_n)\\ &= A\left(\sum\limits_{m=1}^\infty (B e_n,e_m)e_m\right)\\ & = \sum\limits_{m=1}^\infty (B e_n,e_m) A e_m \end{align*} So if we now take the scalar product $(ABe_n,e_n)$ and sum over all $n\in\mathbb{N}$, we get the desired equality: \begin{align*} \sum\limits_{n=1}^\infty (ABe_n,e_n) &= \sum\limits_{n=1}^\infty \left(A \left( \sum\limits_{m=1}^\infty (B e_n,e_m)e_m\right),e_n\right)\\ &= \sum\limits_{n=1}^\infty \sum\limits_{m=1}^\infty (B e_n,e_m) (Ae_m,e_n) \end{align*}

Answer 2

$$B e_n = \sum_{m} \langle B e_n, e_m \rangle e_m \text{, so } \langle A B e_n, e_n \rangle = \sum_m \langle B e_n, e_m \rangle \langle A e_m, e_n \rangle$$ and summing over $n$ yields the result.

Answer 3

For two elements $x, y$ in $H$ by the Parseval identity we have $$ \langle x, y \rangle = \sum_{m = 1}^\infty \langle x, e_m \rangle \overline{\langle y, e_m \rangle} \\ = \sum_{m = 1}^\infty \langle x, e_m \rangle \langle e_m, y \rangle. $$ Hence $$ \langle A B e_n, e_n \rangle = \langle B e_n, A^\ast e_n \rangle \\ = \sum_{m = 1}^\infty \langle B e_n, e_m \rangle \langle e_m, A^\ast e_n \rangle \\ = \sum_{m = 1}^\infty \langle B e_n, e_m \rangle \langle A e_m, e_n \rangle. $$