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I know that the sequence space $l^\infty$ is equal to the dual of $l^1$ with respect to the $\|\cdot\|_1$ norm. But do we also have, that $$ l^\infty = \{f \in \mathbb R^{\mathbb N} \mid \sum_{k=1}^\infty |f_k g_k| < \infty \text{ for all } g \in l^1\} ? $$ Every element in the set on the right hand side obviously defines a linear functional on $l^1$, but is it necessarily continuous? My guess is that the answer is no and there is a connection to $(l^\infty)' \neq l^1$, but I'm not sure how to prove it.

Martin Sleziak
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BeABee
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  • Do you think the map $g \mapsto \sum_{k=1}^\infty |f_k g_k|$ is bounded for each $f \in \mathbb R^{\mathbb N}$? – cejvan Jan 13 '16 at 11:20
  • no. Phrased differently I mean the following: does $\sum_{k=1}^\infty |f_k g_k|< \infty$ for all $f \in l^1$, imply that $|g|_\infty < \infty$? – BeABee Jan 13 '16 at 11:28
  • I think a construction similar to http://math.stackexchange.com/a/674068/58577 should be possible. So the answer should be 'yes'. – gerw Jan 13 '16 at 11:40

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The equality you ask about holds, and there is an elementary proof (of course, only one inclusion has to be shown):

Let $(f_n)$ be an unbounded sequence in $\mathbb{R}$. Then there is subsequence $(f_{n_k})$ such that $|f_{n_k}|\geq 2^k$. Let $|g_m|=2^{-k}$ for $m=n_k$ and $g_m=0$ otherwise. Then $g\in \ell^1$ and $$ \sum_{m=1}^\infty |f_m g_m|=\sum_{k=1}^\infty |f_{n_k}||g_{n_k}|\geq \sum_{k=1}^\infty 1=\infty. $$

MaoWao
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