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Question: Find all $\alpha\in \mathbb{R}$ such that $A=\{(x_n)_{n=1}^{\infty}\in l_p:\sum_{n=1}^{\infty} n^{\alpha}\cdot |x_n|^{p+1}\leq 1\}$ is totally bounded.

In the question, metric is not given. So, I think the metric is $d(x,y)=\left(\sum_{n=1}^{\infty}|x_n-y_n|^p\right)^{1/p}$.

My Work: $A$, a subset of $l_p$ is totally bounded if $\forall \epsilon>0$, $\exists N\in\mathbb{N}$ such that $\sum_{n=N}^{\infty} |x_n|^p<\epsilon$ for all $x\in A$. By Hölder's inequality, $$\sum_{n=N}^{\infty} |x_n|^p\leq \left(\sum_{n=N}^{\infty} n^{\alpha}\cdot |x_n|^{p+1}\right)^{p/(p+1)}\left(\sum_{n=N}^{\infty}n^{-p\alpha}\right)^{1/(p+1)}\leq \left(\sum_{n=N}^{\infty}n^{-p\alpha}\right)^{1/(p+1)}$$ Therefore, if $\alpha>\frac{1}{p}$, then we can find $N\in\mathbb{N}$ such that $\left(\sum_{n=N}^{\infty}n^{-p\alpha}\right)^{1/(p+1)}<\epsilon$ because $a_k=\sum_{n=k}^{\infty}n^{-p\alpha}\to 0$ when $k\to \infty$. Because of that, I think the answer is real numbers larger than $\frac{1}{p}$. To prove that we need to show that $\alpha\leq \frac{1}{p}$ don't work. I defined a new set $$B=\left\{(x_n)_{n=1}^{\infty}\in l_p:\sum_{n=1}^{\infty} n^{\frac{1}{p}}\cdot |x_n|^{p+1}\leq 1\right\}$$ $B\subseteq A$ because if $x\in B$, then $\sum_{n=1}^{\infty} n^{\alpha}\cdot |x_n|^{p+1}\leq \sum_{n=1}^{\infty} n^{\frac{1}{p}}\cdot |x_n|^{p+1}\leq 1$ and $x\in A$. So, it is enough to show that $B$ is not totally bounded. I could not continue after that part. How can I continue?

lecdabster
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1 Answers1

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If $B$ were a totally bounded set, then it also would be a bounded set. Therefore, it is enough to show that it is not bounded. Let's try to find a sequence $(x_n)_{n=1}^{\infty}$ such that $x_n=cn^{-q}$ for some positive constant $c$. Because we are working on $l_p$, we need $q>\frac{1}{p}$.

We also want that $(x_n)_{n=1}^{\infty}\in B$, therefore, $$c^{p+1}\sum_{n=1}^{\infty} n^{\frac{1}{p}-q(p+1)}\leq 1$$ Because $q>\frac{1}{p}$, we have $q(p+1)-\frac{1}{p}>1$ and $$c\leq \frac{1}{\left[\zeta\left(q(p+1)-\frac{1}{p}\right)\right]^{\frac{1}{p+1}}}$$ We can simply choose $c$ as its upperbound we found. Now $x=(x_n)_{n=1}^{\infty}\in B$ and $$\sum_{n=1}^{\infty} |x_n|^p=\frac{\zeta(qp)}{\left[\zeta\left(q(p+1)-\frac{1}{p}\right)\right]^{\frac{p}{p+1}}}\geq \left[\zeta(qp)\right]^{\frac{1}{p+1}}$$ because $qp<q(p+1)-\frac{1}{p}$ and $\zeta(s)$ decreases when $s>1$. However, RHS is goes to $\infty$ when $q\to \frac{1}{p}^+$. So, $B$ is not bounded, and by that, it is not totally bounded. $A$ is totally bounded if and only if $\alpha>\frac{1}{p}$.

lecdabster
  • 205