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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 ?

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    Yes, continuous linear operators!2012-08-23

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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.

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    @user36772: If two functions are equal at every point of their domain then they are the same. This is a set theoretical issue, as you pointed in your question. Whether or not the fact that a limit is continuous implies convergence in the operator norm, I don't know. You might want to search for it here -- and if you cannot find it, ask a new question.2012-08-23
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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.

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    Actually I meant with (1) that the maps $A$ and $K$, with $K=\sum_{k=1}^{\infty} R_k$ are identical. Hence the confusion. The texts I have read also dealt only with the operator norm on $L(H)$.2012-08-23