This was asked in my exam yesterday. I could not do it.
Let $p\in [1,\infty$]. Let $(a_n)$ be a complex sequence such that the sequence $(a_nx_n)\in l^p$ for all sequences $(x_n)\in l^p$ where $p\in [1,\infty]$
Show that $(a_n)\in l^\infty.$
To show that $ |(a_n)|$ is bounded.
Let $ |(a_n)|$ is not bounded.Then $|a_n|$ contains a monotonically increasing subsequence say $|a_{n_k}|$ where $|a_{n_1}|\ge |a_{n_2}|\ge \dots $
Then for each positive integer $k,\exists n_k$ such that $|a_{n_k}|>n_k$.
Take $x_n=\dfrac{1}{n^2}\implies \sum _{n=1}^\infty \dfrac{1}{n^{2p}}<\infty \forall p\in [1,\infty]$
Now $\sum _{n=1}^\infty |x_{n_k}a_{n_k}|^p=\sum _{n=1}^\infty |x_{n_k}|^p|a_{n_k}|^p>\sum _{n=1}^\infty \dfrac{n_k^p}{n_k^{2p}}=\sum _{n=1}^\infty \dfrac{1}{n_k^{p}}=\infty $ when $p=1$.
Now what about the other $p\in (1,\infty]$?
Help needeed.
Suppose $(a_n)$ is unbounded. Then $(a_n)$ has subsequences $(a_{i_n})$ that grow arbitrarily fast in absolute value. In particular, assume $|a_{i_n}| > e^n$ for all $n$. Note that the sequence $(b_n)$ defined by $b_{i_n} = \frac{1}{n^2}$ and which is zero for all other indices is in $l^p$ for all $p\in[1, \infty)$. But $\sum_{n=1}^{\infty} |a_nb_n|^p > \sum_{n=1}^{\infty} e^{np}/n^{2p}$, which diverges to infinity.