With $\mathcal{L}^2(H) = \left\{T \in \mathcal{L}(H), \quad \|T\|_2 < +\infty\right\}$ where $\|T\|_2 = \sum_{i=0}^{+\infty} \|T(b_i)\|^2$ for any Hilbert basis $B = (b_i)_{i \in \mathbb{N}}$ of $H = l^2(\mathbb{N})$, Let $L$ and $U$ be in $\mathcal{L}^2(H)$ and $B = (b_i)_{i \in \mathbb{N}}$ a Hilbert basis of $H$. Show that the quantity $$\sum_{i=0}^{+\infty} \langle L(b_i), U(b_i) \rangle$$ is finite, independent of the basis $B$ chosen, and defines an inner product on $\mathcal{L}^2(H)$.
With $\mathcal{L}^2(H) = \left\{T \in \mathcal{L}(H), \quad \|T\|_2 < +\infty\right\}$ where $\|T\|_2 = \sum_{i=0}^{+\infty} \|T(b_i)\|^2$ for any Hilbert basis $B = (b_i)_{i \in \mathbb{N}}$ of $H = l^2(\mathbb{N})$,
Let $L$ and $U$ be in $\mathcal{L}^2(H)$ and $B = (b_i)_{i \in \mathbb{N}}$ a Hilbert basis of $H$. Show that the quantity
$$\sum_{i=0}^{+\infty} \langle L(b_i), U(b_i) \rangle$$
is finite, independent of the basis $B$ chosen, and defines an inner product on $\mathcal{L}^2(H)$.