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Prove that any sequence $(x^{(n)})_{n\in\mathbb{N}}\subseteq\ell^1$ such that $\sum_{k=1}^\infty k\lvert x_k^{(n)}\lvert\leq1$ for all $n\in\mathbb{N}$ has a convergent subsequence.

My thoughts on this: Clearly $\lvert x^{(n)}_k\lvert\leq\frac{1}{k}$ uniform in $n$. Therefore the sequence $x_1^{(n)}$ has a convergent subsequence $x_1^{(\tilde{n}_k)}$. Further extracting subsequences for any fixed $N\in\mathbb{N}$ I find $({n_l})_{l\in\mathbb{N}}\subseteq\mathbb{N}$ such that $x_k^{(n_l)}$ converges to some $x_k$ for all $1\leq k\leq N$ as $n\rightarrow\infty$.

If for every $\epsilon>0$ I could find some $N\in\mathbb{N}$ such that $\sum_{k=N+1}^\infty \lvert x_k^{(n)}\lvert<\epsilon$ uniform in $n$ the proof would be complete. But I don't see why this should be true.

Can you give me some hint?

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Note: here is, I believe, the result you wanted to prove initially. The fact that the following operator is compact follows easily from the fact that it is the operator norm limit of finite rank operators.

Consider the bounded linear operator $$ T:\ell^1\longrightarrow \ell^1\qquad (x_k)\longmapsto \left(\frac{x_k}{k}\right). $$ Since $T$ is the operator norm limit of the finite rank operators, $$T_n:(x_k)\longmapsto \left(x_1,\frac{x_2}{2},\ldots,\frac{x_n}{n},0,\ldots\right),$$ it is compact. Indeed, $\|T-T_n\|\leq \frac{1}{n+1}$. So if $B$ denotes the closed unit ball of $\ell^1$, then $T(B)$ is relatively compact in $\ell^1$.

Now just observe that $$ T(B)=\{(y_k)\in\ell^1\;;\; \sum_{k=1}^{+\infty} k\,|y_k|\leq 1\} $$ is the set you are considering. It is easily seen to be closed in $\ell^1$. Indeed, if $y^{(n)}$ is in $T(B)$ and converges to $y$ in $\ell^1$, then it converges pointwise to $y$ a fortiori. So for every $K$ and every $n$, we have $$ \sum_{k=1}^K k\,|y^{(n)}_k|\leq 1\quad \Rightarrow \quad \sum_{k=1}^K k\,|y_k|\leq 1\quad \Rightarrow\quad \sum_{k=1}^{+\infty} k\,|y_k|\leq 1. $$

Conclusion: your set $T(B)$ is a compact subset of $\ell^1$. What you want is just the sequential compactness of the latter: every sequence in $T(B)$ has a convergent subsequence in $T(B)$ (and not only in $\ell^1$).

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I think you may want a proof like this:

By the given condition we know $|x_k^{(n)}|\le1$ for all $n$ and $k$. So for $k=1$, we may pick up a convergent subsequence $\{x_1^{n^1_j}\}_{j\in\mathbb{N}}$. Then based on this subsequence we pick up for $k=2$ a convergent subsequence $\{x_2^{n^2_j}\}_{j\in\mathbb{N}}$. Keep doing in this way and get a sequence for each $k$, and then take the diagonal elements $n^j_j$. It can be verified that $\{x^{n^j_j}\}_{j\in\mathbb{N}}$ converge pointwisely to some $\tilde{x}$. For simplicity we denote this new sequence $\{x^{(m)}\}_{m\in\mathbb{N}}$. It can be shown that $\sum_{k=1}^{\infty}k|\tilde{x}_k|\le1$, hence $\tilde{x}\in\mathcal{l}^1$. In fact, since $\{x^{(m)}\}_{m\in\mathbb{N}}$ converge pointwisely to $\tilde{x}$, for finite $N$ we have $\sum_{k=1}^{N}k|\tilde{x}_k|\le1$, then passing $N$ to infinity we know $\sum_{k=1}^{\infty}k|\tilde{x}_k|\le1$.

Then we show that for every $\epsilon>0$ we can find some $N\in\mathbb{N}$ such that $\underset{m}{\sup}\sum_{k=N+1}^{\infty}|x_k^{(m)}|<\epsilon$. In fact, only need to observe that $$\sum_{k=N+1}^{\infty}|x_k^{(m)}|\le\frac{1}{N+1}\sum_{k=N+1}^{\infty}k|x_k^{(m)}|\le\frac{1}{N+1},\qquad\forall m.$$

Now we're already to prove your claim. For any $\epsilon>0$ take $N_1$ such that $\underset{m}{\sup}\sum_{k=N_1+1}^{\infty}|x_k^{(m)}|<\epsilon/4$. Note that $\tilde{x}\in\mathcal{l}^1$, so we can find $N_2$ such that $\sum_{k=N_2+1}^{\infty}|\tilde{x}_k|<\epsilon/4$. Therefore, if taking $N_3=\max\{N_1,N_2\}$, then $$\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}-\tilde{x}_k|\le\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}|+\sum_{k=N_3+1}^{\infty}|\tilde{x}_k|<\epsilon/2.$$ On the other hand, since $\{x^{m}\}_{m\in\mathbb{N}}$ converge pointwisely to $\tilde{x}$, we can find $N_4$ such that $$\sum_{k=1}^{N_3}|x_k^{(m)}-\tilde{x}_k|<\epsilon/2,\qquad\forall m>N_4.$$ Combining the above two we know $$\|x^{(m)}-\tilde{x}\|_{\mathcal{l}^1}\le\sum_{k=1}^{N_3}|x_k^{(m)}-\tilde{x}_k|+\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}-\tilde{x}_k|<\epsilon,\qquad\forall m>N_4.$$ This means $\{x^{(m)}\}$ converge to $\tilde{x}$ in $\mathcal{l}^1$. Proof is finished.

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