Let $(x_{n})_{n}$ be a $p$-summable sequence in a Banach space $X$. Define an operator $T$ from $l_{q}$(where $q=p/(p-1)$) to $X$ by $Te_{n}=x_{n}(n=1,2...)$. Is $T$ a $p$-summing operator?
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asked May 8, 2015 at 8:01
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Not unless $p=1$, in which case you interpret $\ell_q$ to be $c_0$.
The most interesting case is $p=2$. Tomczak's theorem says that you can compute $\pi_2(T)$ for $T: \ell_2^n \to X$ up to the constant $\sqrt{2}$ by taking the sup of $(\sum_{j=1}^n \| Tx_j \|^2)^{1/2}$, where the sup is over ALL orthonormal bases of $\ell_2^n$.
To see that you cannot get by with just one orthonormal basis, consider the formal identity $I_{2,1} : L_2^n \to L_1^n$. Here I am using the function space normalization; i.e., the measure on $\{1,\dots,n\}$ is the uniform probability measure rather than counting measure, so that $\|I_{2,1}\| = 1$ and $\|e_k\|_p = n^{-1/p}$, where, as usual, $e_k(j) := \delta_{k,j}$. So $\{\sqrt{n}e_j\}_{j=1}^n$ is orthogonal and $ (\sum_{j=1}^n \| I_{2,1}(\sqrt{n}e_j)\|_1^{2})^{1/2} = 1$. However, if you do the same computation using the [orthonormal] Walsh basis $\{w_j\}_{j=1}^n$ for $L_2^n$ (assume, for simplicity, that $n$ is a power of $2$ so that this makes sense), you see that $\pi_2(I_{2,1} )= \sqrt{n}$.
answered May 9, 2015 at 10:12
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$\begingroup$ As we know, a $p$-summing operator maps a weakly $p$-summable sequence to a $p$-summable sequence. Conversely,given a $p$-summable sequence $(x_{n})_{n}$ in $X$, do there exist a space $Z$,a $p$-summing operator $T$ from $Z$ to $X$ and a weakly $p$-summable sequence $(z_{n})_{n}$ in $Z$ so that $Tz_{n}=x_{n}$ for all $n$. $\endgroup$ Commented May 10, 2015 at 2:02
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