Interchangability of arbitrary sums and linear operator Let $\mathcal H$ be a Hilbert space, $\{x_i : i \in I \}$ be a orthonormal base in $\mathcal H$ and $T \in L(\mathcal H)$. 
Does the following hold:
$T( \sum_{i \in I} \lambda_i x_i) =  \sum_{i \in I} \lambda_i T(x_i)$ ?
For example, consider $\{ e_i : i \in I\}$, the standard base in $\ell^2(I)$.

We have for finite sums that:
$$
T\left(\sum_{ k = 1}^{N} \lambda x_k \right) = \sum_{k = 1}^{N} \lambda_kT(x_k)
$$
Because $T$ is continuous, we have that for some sequence $x_n \in \mathcal{H}$, $x_n \rightarrow x$ that:
$$
T(x_n) \rightarrow T(x)
$$
Hence, becuase:
$$
\sum_{ k = 1}^{\infty} \lambda_k x_k= \lim_{N \to \infty}\sum_{ k = 1}^{N} \lambda_k x_k
$$
We conclude that:
$$
\lim_{N \to \infty}T\left(\sum_{k = 1}^{N} \lambda_k x_k \right) = \lim_{N \to \infty}\sum_{k = 1}^{N} \lambda_kT(x_k)
$$
The left hand side (by continuity) is:
$$
T\left(\sum_{k = 1}^{\infty} \lambda_k x_k\right)
$$
And the right hand side (by definition) is:
$$
\sum_{k = 1}^{\infty} T(\lambda_k x_k)
$$
as we need.
