p−summing operators space is a Banach space (proof) 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.
 A: 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.
