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Let $X$ be a Banach space and let $T_n\colon X\to X$ be a family of bounded operators convergent to some operator $T\colon X\to X$. Is it true that

$T(X)\subseteq \sum_{n=1}^\infty T_n(X)$?

I mean by $\sum_{n=1}^\infty V_n$ the set of all (finite) sums of the form $v_{i_1}+\ldots+v_{i_n}$ where $v_{i_k}\in V_{i_k}$ and $V_{i_k}\subseteq X$.

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I have edited the post after mae's answer. Sorry. –  dziobak Jun 11 '11 at 23:08

1 Answer 1

No. Take $T_n:\ell^2\to \ell^2$, $(x_1,x_2,\dots)\mapsto (x_1,x_2/2,\dots,x_n/n,0,0,0,\dots)$. Every vector in $T_n(\ell^2)$ has finite support, hence any vector in $\sum_{n=1}^\infty T_n(X)$ also has finite support. But the range of $T=\lim T_n$ contains vectors of full support, such as $T(1,1/2,1/3,1/4,\dots)=(1,1/4,1/9,1/16,\dots)$.

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Sorry, I meant the algebraic sum (see the correction). –  dziobak Jun 11 '11 at 23:07
    
Well, since you're only considering finite sums, I think the same counterexample works. –  mac Jun 11 '11 at 23:08
    
How abut the closure of this set? –  dziobak Jun 11 '11 at 23:10
3  
Ah, the closure should work. I guess that if $T_n\to T$ then $T_n(x)\to T(x)$ for every $x$, so $T(x)$ is always in the closure of $\bigcup_n T_n(X)$ (and this is contained in the closure of $\sum_n T_n(X)$). –  mac Jun 11 '11 at 23:13
    
@dziobak: Why do you need the answer to this question? Do you use it somewhere? You changed the question two times already. Do you want to get somewhere with this? –  Beni Bogosel Jun 12 '11 at 13:53

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