# $p$-summing operators space is a Banach space

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.

• I use the terminology by Lindenstrauss-Jafriri/ Classical Banach Spaces I seqence spaces book – Kostas Apr 29 at 14:36
• 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 – Yemon Choi Apr 29 at 15:15
• 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. – Kostas Apr 29 at 16:13
• 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 – Kostas Apr 29 at 21:13
• 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. – Bill Johnson Apr 29 at 21:39