Is $\|\cdot\|_{\pi} = \|\cdot\|$? Let $\ell^{1}(\mathbb{N})$ be the space of complex-valued sequences such as:
$$\|a\|_{\ell^{1}} = \sum_{n\in \mathbb{N}}|a_{n}| < +\infty$$
and set $\ell^{1}(\mathbb{R})\otimes \ell^{1}(\mathbb{R})$ to be the set of all finite sums of sequences of the form $a\otimes b = \{(a\otimes b)_{n,m}\}_{n,m \in \mathbb{N}}$, with $(a\otimes b)_{n,m} = a_{n}b_{m}$. If $c = \sum_{k}a_{k}\otimes b_{k}$ is an element of this space, define:
$$\|c\| := \sum_{n,m}|\sum_{k}(a_{k}\otimes b_{k})_{n,m}|$$
In addition, define:
$$\|c\|_{\pi} := \inf \sum_{k}\|a\|_{\ell^{1}}\|b\|_{\ell^{1}}$$
where the infimum is over all possible representations of $c$ as a sum of finite elements of $a_{k}\otimes b_{k}$.
Do $\|\cdot\|$ and $\|\cdot\|_{\pi}$ agree? I proved $\|\cdot\| \le \|\cdot\|_{\pi}$ but not the converse. I believe the $\|\cdot\|_{\pi}$ is called projective norm in the literature.
 A: I will use the convention where $0\in\mathbb{N}$.
Let $N\ge 0$ be large enough such that $n>N$ or $m>N$ implies $c_{n,m}=0$.
Let $\sigma,\tau$ be maps from $\{0,\ldots, (N+1)^2-1\}$ to $\{0,1,\ldots,N\}$ such that $k\mapsto(\sigma(k),\tau(k))$ is a bijection from  $\{0,\ldots, (N+1)^2-1\}$ onto $\{0,1,\ldots,N\}^2$.
For $0\le k\le N^2+2N$, and using Kronecker deltas, define $a_k\in\ell^1$ by
$$
(a_k)_n=c_{\sigma(k),\tau(k)}\ \delta_{n,\sigma(k)}\ .
$$
Likewise define $a_k\in\ell^1$ by
$$
(b_k)_m=\delta_{m,\tau(k)}\ .
$$
Then it is easy to see that $c=\sum_{k=0}^{N^2+2N}a_k\otimes b_k$.
Moreover for all $k$, we have $\|a_k\|_{\ell^1}=|c_{\sigma(k),\tau(k)}|$ and $\|a_k\|_{\ell^1}=1$. So
$$
\|c\|_{\pi}\le \sum_{k=0}^{N^2+2N}\|a_k\|_{\ell^1}\ \|b_k\|_{\ell^1}=\sum_{k=0}^{N^2+2N}|c_{\sigma(k),\tau(k)}|=\|c\|\ .
$$
Remark:
I briefly hinted at the solution above after Proposition 2 in my MO answer
https://mathoverflow.net/questions/363935/what-is-the-role-of-topology-on-infinite-dimensional-exterior-algebras/364211#364211
