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Given $p\in (1,\infty)$. Take a bounded sequence $(f_n)$ in $\ell^p$ and define a linear map $T\colon \ell^p_{00}\to \ell^p$ ($\ell^p_{00}$ is the space of finitelty supported sequences in $\ell^p$) by

$Te_n = f_n$

($e_n$ is the usual (Hamel) basis for $\ell^p_{00}$).

Is $T$ a bounded linear map? If $p=1$, the answer is yes... If not, what do have to assume on $(f_n)$ to get boundeness of $T$?

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In $\ell^2_{00}$ try $f_n = e_1$ for all $n$. So you do need some assumptions on $f_n$. Even in $\ell_2$ I think there is no simple criterion. "Simple" meaning easier than writing down the definition for a bounded operator. –  GEdgar May 24 '12 at 12:19
    
Edgar's example should work in all $\ell^p_{00}$ (as for no $p > 1$ it holds that $n = \|ne_1\|_p \le Cn^{1/p} = C\|(1,\ldots, 1,0,\ldots)\|_p$ is true for all $n$). –  martini May 24 '12 at 12:26
    
$T$ will for example (to give a sufficent criterion) be bounded if the $f_n$ are disjointly supported as then $\|T(\sum_i x_i e_i)\|_p^p = \|\sum_i x_i f_i\|_p^p = \sum_i \|x_i f_i\|_p^p \le M^p\sum_i |x_i|^p = M^p\|x\|_p^p$ for all $x \in \ell^p_{00}$ where $M$ is some upper bound of the $\|f_i\|_p$. –  martini May 24 '12 at 12:30
    
One nontrivial sufficient condition is the Schur Test ($p=2$), which you can find, for example, in Conway's Course in Functional Analysis. [The Wikipedia article states it in the integral form only.] –  user31373 May 24 '12 at 14:51

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