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Let X,Y Banach spaces and $p\geq 1$. A bounded linear operator $T$ is called p-absolutely summing, if there is exist $K>0$, such that for all $n\in N$ and $x_1,\dots, x_n\in X$: \begin{equation}\label{p-summing-def} \left(\sum_{i=1}^n \|T(x_i)\|^p\right)^{1/p}\leq K\cdot\sup_{\|x^*\|\leq1}\left(\sum_{i=1}^n \|x^*(x_i)\|^p\right)^{1/p}. \end{equation} The smallest Κ such that the previus condition holds denoted by $\pi_p(T)$. Also, the set of all p-absolutly summing operators denoted by $\Pi_p(X,Y)$. When an operator $T\not\in \Pi_p(X,Y)$, we write $\pi_p(T)=+\infty$. Easily, we can proof that $\Pi_p(X,Y)$ is a normed space with $\pi_p(\cdot)$ as a norm. I stuck in the middle to proof that $\Pi_p(X,Y)$ with the norm $\pi_p(\cdot) $ is a Banach space. My idea is to show that every $(T_n)$ Cauchy sequence in $\Pi_p(X,Y)$ is also convergent in $\Pi_p(X,Y)$. By hypothesis we get that forall $x\in X$, $(T_n(x))$ converges at $T(x)$, for some T. Then I have the idea to write $T=S+P$, where S and P are p-absolutlly summing operators (like in the proof to showing that the space of linear bounded operators $\mathcal{B}(X,Y)$ is a Banach space, when Y is a Banach space). I believe that $S= T_{n_0}$ and $P=T-T_{n_0}$, and we get $T_{n_0}$ from the Cauchy Hypothesis and convegence in Y, but I am not sure about this, is just a guess. Can you help me to complete the proof or at least give me some ideas. Thank you.

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    $\begingroup$ You should define what the notation in this question means. What is the space $\Pi_p(X,Y)$ and what is its norm $\pi_p$? $\endgroup$ Apr 29, 2019 at 16:24
  • $\begingroup$ I use the terminology by Lindenstrauss-Jafriri/ Classical Banach Spaces I seqence spaces book $\endgroup$
    – Kostas
    Apr 29, 2019 at 20:31
  • $\begingroup$ I and many other people on this site don't have access to that book. You should try to make your question self-contained by editing it to define the objects involved. $\endgroup$ Apr 29, 2019 at 20:37
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    $\begingroup$ I edited it. Much better I guess $\endgroup$
    – Kostas
    Apr 29, 2019 at 21:07

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You've identified that if $T_n$ is a Cauchy sequence for $\pi_p$ then it converges pointwise to some linear map $T$. Further, since $$\|Tx\| = \lim_{n \to \infty} \|T_nx\| \leq \sup_n \pi_p(T_n) \cdot \sup_{\|x^*\| \leq 1} |x^*(x)| = \sup_n \pi_p(T_n) \|x\|$$ $T$ is a bounded operator.

It remains to check that $\pi_p(T) < \infty$ and $\pi_p(T_n-T) \to 0$ as $n \to \infty$. Fix $x_1, \dots, x_n \in X$ and pick $N$ large enough that $$\bigg(\sum_{i=1}^n \|Tx_i - T_Nx_i\|^p \bigg)^{\frac{1}{p}} \leq \sup_{\|x^*\| \leq 1} \bigg (\sum_{i=1}^n|x^*(x_i)|^p \bigg)^{\frac{1}{p}}$$ (possible by pointwise convergence). Then we can estimate \begin{align*} \bigg( \sum_{i=1}^n \|Tx_i\|^p \bigg)^{\frac{1}{p}} \leq& \bigg( \sum_{i=1}^n \|T_Nx_i\|^p \bigg)^{\frac{1}{p}} + \bigg( \sum_{i=1}^n \|Tx_i - T_Nx_i\|^p \bigg)^{\frac{1}{p}} \\ \leq& \pi_p(T_N)\sup_{\|x^*\| \leq 1} \bigg (\sum_{i=1}^n |x^*(x_i)|^p \bigg)^{\frac{1}{p}} + \sup_{\|x^*\| \leq 1} \bigg (\sum_{i=1}^n |x^*(x_i)|^p \bigg)^{\frac{1}{p}} \\ \leq& [1+ \sup_n \pi_p(T_n)]\sup_{\|x^*\| \leq 1} \bigg (\sum_{i=1}^n |x^*(x_i)|^p \bigg)^{\frac{1}{p}} \end{align*} so $\pi_p(T) \leq 1 + \sup_n \pi_p(T_n) < \infty$.

Now we want to prove that $\pi_p(T_n - T) \to 0$. Note that for all $\varepsilon > 0$ there is an $N$ such that $n,m \geq N$ implies that $\pi_p(T_n-T_m) < \varepsilon$. Then, for any $x_1, \dots x_k$ and $n \geq N$, $$\bigg( \sum_{i=1}^k \|Tx_i - T_nx_i\|^p \bigg)^{\frac{1}{p}} = \lim_{m \to \infty} \bigg( \sum_{i=1}^k \|T_m x_i - T_nx_i\|^p \bigg)^{\frac{1}{p}} \leq \varepsilon \sup_{\|x^*\| \leq 1} \bigg (\sum_{i=1}^k |x^*(x_i)|^p \bigg)^{\frac{1}{p}}$$ and so for $n \geq N$, $\pi_p(T_n - T) \leq \varepsilon$ which gives the desired convergence.

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  • $\begingroup$ The template for this proof is a very generic one. To check a function space is complete, we often show that Cauchy sequences must converge pointwise, then show that the pointwise limit lies in the space we are considering and finally show convergence in the actual norm. See, for example, the usual proofs that $\ell^p$ or $C([0,1])$ are complete. For another "less usual" function space where I use the exact same template, see my answer to a question asking about a space of vector measures with the total variation norm here. $\endgroup$ Apr 29, 2019 at 21:54
  • $\begingroup$ Thank you very much I got your idea and that's the point i missed. One more question. In the middle you pick an N large enought. What's the criterion you picked it? I think you pick the N because the $(T_n)$ is a Cauchy sequence and then? $\endgroup$
    – Kostas
    May 1, 2019 at 10:11
  • $\begingroup$ You mean where I say "Fix $x_1, \dots, x_n \in X$ and pick $N$ large enough that..."? Well notice that if the right hand side in the following inequality is $0$ then so is each $x_i$ so the inequality is trivial. So we can assume the right hand side is positive. Then $T_kx_i \to Tx_i$ for each $i$ as $k \to \infty$ by definition. This implies that $\bigg(\sum_{i=1}^n \|Tx_i - T_kx_i\|^p \bigg)^{\frac{1}{p}} \to 0$ as $k \to \infty$. We then use this to choose $N$ large enough that the desired inequality holds. $\endgroup$ May 1, 2019 at 14:31
  • $\begingroup$ Perfect!! I understood completely the proof. Thank you very much for your help $\endgroup$
    – Kostas
    May 1, 2019 at 16:18
  • $\begingroup$ No problem! If you're happy with it, you might consider clicking the tick you see by my answer so the question is marked as answered on the site $\endgroup$ May 1, 2019 at 18:57

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