Let $X,Y$ be 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 $x_1,\dots, x_n\in X$: $$ \left(\sum_{i=1}^n \|T(x_i)\|^p\right)^{1/p}\leq K\cdot\sup_{\|x^*\|\leq1}\left(\sum_{i=1}^n \|x^*(x_i)\|^p\right)^{1/p}. $$ The smallest $Κ$ such that the previous condition holds is denoted by $\pi_p(T)$. Also, the set of all p-absolutely summing operators is 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 prove that $\Pi_p(X,Y)$ is a normed space with $\pi_p(\cdot)$ as a norm. I am stuck in the middle of the 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 for all $x\in X$, $(T_n(x))$ converges at $T(x)$, for some T. Then I have the idea to write $T=S+P$, where $S$ and $P$ are $p$-absolutely 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 convergence in Y, but I am not sure about this, is just an guess. Can you help me to complete the proof or at least give me some ideas. Thank you.

  • $\begingroup$ I use the terminology by Lindenstrauss-Jafriri/ Classical Banach Spaces I seqence spaces book $\endgroup$ – Kostas Apr 29 at 14:36
  • 1
    $\begingroup$ Take an $m$-tuple of test elements $(x_1,\dots, x_m)$ and use $\Vert Tx_j\Vert = \lim_{n\to \infty} \Vert T_nx_j\Vert$. I think this hint should be sufficient; if you need further details, then I think math.stackexchange.com is a more suitable venue for this kind of question, especially if you are following a textbook $\endgroup$ – Yemon Choi Apr 29 at 15:15
  • $\begingroup$ Ok thank you. I think I did it! I pass the limit at the inequality that $(T_n)$ satisfies (because is a sequence of p-absolutlly summing operators) and then I have the same inequality satisfies but this time for the operator T. So, T is p-absolutelly summing operator. $\endgroup$ – Kostas Apr 29 at 16:13
  • $\begingroup$ I didn't do it :( because the constant K depents from n so I have a $(K_n) $ sequense of positive real numbers from hypothesis that I don't know whether converges $\endgroup$ – Kostas Apr 29 at 21:13
  • 1
    $\begingroup$ You can look it up in books (e.g., Diestel-Jarchow-Tonge), but here is a hint that is useful when checking that a normed space is complete. $\endgroup$ – Bill Johnson Apr 29 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.