Absolutely 2-summable operator on a Hilbert space

Question

An bouneded linear operator $A \in L(X, Y)$ (here $X$, $Y$ are Banach spaces) is called absolutely $2$-summable if there exists a $C>0$ such that $$ \left( \sum_{j=1}^N \| A x_j\|_X^2 \right)^{1/2} \leq C \cdot \sup \left\{ \left(\sum | \langle x_j, \omega \rangle | \right)^{1/2} \mid \omega \in X', \|\omega\|_{X'} = 1 \right\}$$ for any finite set of vectors $x_1, \dots x_N \in X$. The $2$-summable norm is the infimum of all such constants $C$.

Those operators are the natural analog to Hilbert-Schmidt-Operators: If $X$ and $Y$ are Hilbert spaces, than $A$ is absolutely $2$-summable if and only if $A$ is Hilbert-Schmidt, and the norms coincide.

Now the question is: If $X$ is a Hilbert space (but $Y$ isn't), can we restrict to elements $x_j$ of some orthonormal basis? That is, if we allow only elements of a given ONB to take for $x_1, \dots, x_N$, can the resulting $C$ be strictly smaller than the $2$-summable norm?

Answer 1

I answered the OP's question in a comment, but the right question is whether you can estimate the $2$-summing norm of an operator $T$ from a Hilbert space $H$ by taking $\sup (\sum \|Te_n\|^2)^{1/2}$, where the sup is over all ON bases $(e_n)$ for $H$. In fact, if $H$ is infinite dimensional, this sup is equal to the $2$-summing norm (combine a dilation argument with the fact that $T$ is essentially zero on an infinite dimensional subspace). More interesting is that the the sup is at least $2^{-1/2}$ times the $2$-summing norm of $T$. This is the gist of Tomczak's lemma (see Theorem 18.4 in her book "Banach-Mazur distances and finite-dimensional operator ideals").