Let $A$ be a Banach space and $\mathcal{H}$ be a Hilbert space. Let $x: A\to\mathcal{H}$ be a bounded linear map. A nuclear decomposition is a family of linear functionals $\phi_k \in A^*$ and vectors $\Phi_k\in\mathcal{H}$ such that $x = \sum_k \phi_k(\cdot)\Phi_k$. The nuclear norm of $x$ is denoted as $\|x\|_1$ and is defined as the infimum of $\sum_k\|\phi_k\|\|\Phi_k\|$ for all such nuclear decompositions, and if it is finite, then $x$ is said to be nuclear.
QUESTION 1. For $\epsilon>0$, can we find a nuclear decomposition $x = \sum_k \phi_k(\cdot)\Phi_k$ such that the vectors $\Phi_k$ form an orthonormal system, and that $\sum_k \|\phi_k\|\|\Phi_k\| \le \|x\|_1+\epsilon$?
QUESTION 2. (weaker) For $\epsilon>0$, is there a $C>0$ such that we can find a nuclear decomposition $x = \sum_k \phi_k(\cdot)\Phi_k$ with the vectors $\Phi_k$ forming an orthonormal system, and such that $\sum_k \|\phi_k\|\|\Phi_k\| \le C\|x\|_1+\epsilon$?
Commentary. By definition, if $x$ is nuclear and $\|x\|_1$ is its nuclear norm, then for any $\epsilon >0$ there is a nuclear decomposition $x = \sum_k \phi_k(\cdot)\Phi_k$ such that $\sum_k \|\phi_k\|\|\Phi_k\| \le \|x\|_1+\epsilon$. The family of vectors $\{\Phi_k\}_k$ needn't to be orthogonal, though. A Gram-Schmidt procedure yields a positive answer for question 2 when $\mathcal{H}$ is finite dimensional, with $C = \sqrt{\dim\mathcal{H}}$. I wonder what kind of argument can be given in general cases.