# Strongly convergent series of bounded self-adjoint operators

Let $A_n$ and $A$ be bounded self-adjoint operators in a Hilbert space, such that $A_n\to A$ strongly. Then it is well known that $(z-A_n)^{-1}\to(z-A)^{-1}$ strongly for each $z\in\mathbb{C}\setminus\mathbb{R}$, and even strong convergence of $f(A_n)$ to $f(A)$ for nice enough functions $f$.

My question is a bit similar to this, but involves series, instead of sequences. Let $T_n$ and $T$ be bounded self-adjoint operators, such that $$T_1+\ldots+T_n\to T \qquad \textrm{strongly}.$$ Is it true that $$\frac{T_1}{z-T_1} + \ldots + \frac{T_n}{z-T_n} ,$$ converges strongly to some self-adjoint operator? Do we have convergence of $$f(T_1)T_1+\ldots+f(T_n)T_n ,$$ for nice enough functions? A particular case of interest is $f$ being the indicator function of the positive half line. If it helps, $T$ can be assumed to be invertible.

Regarding the aforementioned expectations:

• Are they hopelessly false?
• If it is salvageable, what kind of extra conditions does one need?
• What is the correct keyword to look up?

Hopelessly false. Consider the one-dimensional case (so the $T_j$ are just numbers). Any function $f$ for which $\sum_j T_j$ convergent implies $\sum_j f(T_j)$ convergent is linear in a neighbourhood of $0$.
Let $f: X \to Y$ be a mapping of normed spaces such that $\sum_{n=1}^\infty f(a_n)$ converges whenever $\sum_{n=1}^\infty a_n$ converges (both in the norm topology). Then there is a neighbourhood of 0 on which $f$ is equal to a bounded linear operator.
Let $f: X \to Y$ be a mapping of Banach spaces such that $\sum_{n=1}^\infty f(a_n)$ converges weakly whenever $\sum_{n=1}^\infty a_n$ converges (strongly). Then there is a neighbourhood of 0 on which $f$ is equal to a bounded linear operator.
• Thank you for the answer! So $f(x)=\max\{0,x\}$ seems to be also OK. My main interest is in this function. I am still hoping.. Aug 20, 2014 at 3:48
• No, it is not OK. A neighbourhood of $0$ includes both sides. Aug 20, 2014 at 3:53