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.

  • $\begingroup$ Don't get why you're starting with $\sum |a_n|<\infty.$ That's $l^1,$ not $l^\infty.$ $\endgroup$ – spaceisdarkgreen Jan 20 '17 at 5:26
  • $\begingroup$ Yes you are right I missed it completely;Can you give some hints to solve it $\endgroup$ – Learnmore Jan 20 '17 at 5:34
  • $\begingroup$ I have added more efforts @spaceisdarkgreen $\endgroup$ – Learnmore Jan 20 '17 at 6:09
  • $\begingroup$ Your argument for the $p = 1$ case has an error, by the way: you seem to have assumed that the sum over any subsequence of $(a_n = 1/n)$ diverges, which is not the case. Also, I think you meant the index of summation to be $k$ in your last few sums, rather than $n$. $\endgroup$ – Vik78 Jan 20 '17 at 6:20

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.

  • $\begingroup$ $x_n=n$ is unbounded but is $n>e^n$?? $\endgroup$ – Learnmore Jan 20 '17 at 6:08
  • $\begingroup$ Please do not downvote if you want me to clarify. $x_n= n$ certainly has a subsequence $(x_{i_n})$ with $x_{i_n}> e^n$: for example, take $i_n = 4^n$. $\endgroup$ – Vik78 Jan 20 '17 at 6:12
  • $\begingroup$ I got your point ; you meant a subsequence not the original sequence; One thing Is the index in the last line $n$ should not it be $i_n$? $\endgroup$ – Learnmore Jan 20 '17 at 6:20
  • $\begingroup$ No, it shouldn't, because we are still summing over all $n$ in the second to last sum. $\endgroup$ – Vik78 Jan 20 '17 at 16:23

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