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I need to understand is the set $\{x\in \ell_1 :\sum_{n=1}^{\infty} \frac{\lvert x_n \rvert}{\sin\left(\frac{1}{n}\right)} \leq 1 \}$ precompact in $\ell_1$.

I tryed to use criterion: set in $\ell_1$ is precompact if and only if it is bounded and $\forall \varepsilon > 0 \exists N \in \mathbb{N}:\forall x \sum_{n=N+1}^{\infty} \lvert x_n \rvert < \varepsilon$.

Boundness is obvious.

I had idea to use $\lvert x_n \rvert \leq \sin\left(\frac{1}{n}\right) $ but i think it is useless because $ \sum_{n=1}^{\infty}\sin\left(\frac{1}{n}\right)$ diverges.

Is it precompact or not?

Denis
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2 Answers2

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For a sequence $a_n>0$ and $a_n\to 0$ consider the operator $$Tx=\sum_{k=1}^\infty a_kx_ke_k$$ where $\{e_k\}$ denotes the standard basis in $\ell^1.$ The operator $T$ is compact as the norm limit of finite dimensional operators $$T_nx=\sum_{k=1}^na_kx_ke_k$$ Therefore $T(B_1)$ is a precompact set, where $B_1$ denotes the unit ball. However $$T(B_1)=\left \{y\in \ell^1\,:\,\sum_{k=1}^\infty a_k^{-1}|y_k|\leq 1\right \}$$ We can apply the above to $a_n=\sin(n^{-1}).$ The same reasoning can be performed for any $\ell^p$ space, $1\le p\le \infty.$

Ryszard Szwarc
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Since $\sin (x) \le x$, we have that ${n} \le \frac{1}{\sin(1/n)} $ for $n > 0$.

Let $\varepsilon > 0$ be given. We have by Cauchy's inequality that for any $N > 1$ $$ \begin{align*} \sum_{n=N}^{\infty}|x_n| &= \sum_{n=N}^{\infty}n\cdot\frac{1}{n}|x_n|\\ &\le \left(\sum_{n=N}^\infty\frac{1}{n^2} \right)^{1/2}\left( \sum_{n=N}^\infty n^2 |x_n|^2 \right)^{1/2} \\ &\le \left(\sum_{n=N}^\infty\frac{1}{n^2} \right)^{1/2}\left( \sum_{n=N}^\infty n|x_n| \right)\\ &\le \varepsilon \sum_{n=1}^\infty \frac{|x_n|}{\sin(1/n)} \le \varepsilon \end{align*} $$ for $N$ chosen large enough and independent of $(x_n)$. We have used the inclusion of the $\ell_p$ spaces (i.e. $\Vert x \Vert_2 \le \Vert x \Vert_1$) in passing to the last line. We have shown that the given set satisfies your criterion.

fwd
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