Confusion about pointwise vs. uniform convergence I'm confused about expressions like $$L=\sum_{k=1}^{\infty} R_k, \quad \quad (1)$$where $L:H\rightarrow H$ and $R_k:H \rightarrow H$ are continuous linear operators in a Hilbert space $H$. We assume $\sum_{k=1}^{\infty} R_k x$ is convergent every $x\in H$.
The thing I'm confused about, is: If we know that for  every $x\in H$ we have $$Lx=\sum_{k=1}^{\infty} R_k x,$$can we that automatically conclude that $(1)$ holds ? 
I would conjecture "yes" since two maps are identical if they have the same domain and range (trivial here) and are identical for every argument - which they are, as the above shows.
But if the answer is indeed "yes", why is it in sum texts shown, that the $R_k$ additionally converge in the operator norm to $L$, i.e. $$\left\|L- \sum_{k=1}^{n} R_k\right\| \rightarrow 0\quad (n\rightarrow \infty),$$ before they conclude that $(1)$ is true ? Was that really necessary ?
 A: It depends on how expression (1) is to be interpreted. The meanig of (1) is that the sequence $S_n = \sum_{k=1}^n R_k$ of operators converges to $L$, right? But $L(H)$ can be used with different topologies. First the norm topology, then for (1) to hold we need to have 
$$ \left\| L - S_n\right\| \to 0, $$
then for example, the strong operator topology, then for (1) to hold we need 
$$ S_nx \to Lx, \quad x \in H $$
or the weak operator topology, then (1) means 
$$ \left< S_n x, y\right> \to \left< Lx, y \right>, \quad x,y \in H. $$
So watch out in the text near (1) which topology is meant.
A: One of the issues with convergence is that it need not preserve continuity.
The simplest example (although not too relevant here) is $x^n$ on $[0,1]$ converging to a discontinuous function.
Another example is convergence for a series of rationals in $\mathbb Q$. If the limit is irrational, then it does not exist (although it does exist in a larger space). This case is similar, if the limit operator is not continuous then it does not exist in our space -- and the sequence is not convergent.
Pointwise convergence is indeed enough to show that $L$ is well-defined, but it is not enough to show that it is continuous. If we wish to talk about convergence in a space of continuous linear operators then we must also verify that the convergence is strong enough for that, namely that the limit operator is itself continuous.
