# A criterion for convergence in the operator norm

Let $L:H\rightarrow H$ be a continuous linear operator and $R_n:H \rightarrow H$ a sequence of continuous linear operators, where $H$ is a Hilbert space. If the $\sum_{n=1}^{k} R_n$ converge pointwise to $L$, meaning for every $x\in H$ we have $$Lx=\sum_{k=1}^{\infty} R_k x=\lim_{k\rightarrow \infty}\sum_{n=1}^{k} R_nx,$$ is then do we know then, that the operator $T$ defined by $$Tx=\sum_{k=1}^{\infty} R_kx,$$

1) equals L ? 2) is continuous ? 3) convergences in the operator norm to $L$ ?

( For 1) and 2) with the help of Asaf Karagila, I think the answer is "yes", since for 1) $L$ and $T$ agree for every value and for 2) we know that $L$ was continuous, so using 1), also $T$ has to be continuous, right? )

EDIT: Everything has been cleared except that I need a counterexample for 3)

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You don't have to assume that $L$ is continuous, it follows automatically from the uniform boundedness principle. – Ahriman Aug 24 '12 at 9:20
What do you mean with you question 3) ? Do you want to know if $\sum_{k=1}^n R_k$ converges in operator norm to $L$ ? This in not necessarily true, there are surely counterexamples in textbooks. – Ahriman Aug 24 '12 at 9:22
@Ahriman Yes, that's what I meant. I just thought that would be true, because convergece in the operator norm would imply continuity of $L$, so I thought that maybe the "converse" would hold, i.e. if $L$ were continuous, it would converge to it in the operator norm. (cont) – user10324 Aug 24 '12 at 10:24
(cont) So 1) and 2) are true ? A counterexample by the would also be a (negative) answer... – user10324 Aug 24 '12 at 10:25
Yes 1) and 2) are true, I don't even see where there could be a problem. – Ahriman Aug 24 '12 at 10:49

Here is a simple counterexample for 3) :

Suppose your Hilbert space $H$ is infinite dimensional and separable, meaning it has an infinite countable orthonormal complete system.

Let denote such a system by $(e_n)_{n \ge 0}$. We know, from Parseval's identity, that every $x \in H$ is equal to $\displaystyle \sum_{n \ge 0} \langle x, e_n \rangle e_n$.

Let $R_n$ defined by $R_n x = \langle x, e_n \rangle e_n$. Then $\sum_{k=0}^n R_k$ converge pointwise to the identity operator.

But if the convergence occurs for the operator norm, since every operator $\sum_{k=0}^n R_k$ is of finite rank, the limit would be a compact operator. But the identity is not compact by Riesz theorem, since $H$ is not finite dimensional.

The special case $H = \ell^2(\mathbb{N})$ with the complete orthonormal system given by $e_n(k) = \delta_{n,k}$ doesn't even require the use of Parseval's identity.

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