I have the following question:
I started off as follows: Let $v\in \ell(\mathbb{N}, \mathbb{C})$, and $w,x$ defined as above. Then: \begin{align} \langle w, x\rangle_J &= \sum_{(i,j,k)\in J}w_{[i, j, k]} \bar{x}_{[i,j,k]}\\ &=\sum_{(i,j,k)\in J} |v_{[j]}||v_{[k]}| && \text{definition of $w,x$}\\ \end{align} However I have literally no idea about how to go forward..
Ideas
My main ideas are that supposedly the matrix $A$ will have at most $N\times N$ non-zero entries. Also, the definition of $J$ is such that the first index of each triplet $(i,j,k)$ basically represents one of $N$ rows containing at most $N$ elements, and the indexes $j$ and $k$ represents one of $N$ columns having at most $N$ elements.
Observe the following estimate:$$\begin{eqnarray} \langle w, x\rangle_J &=& \sum_{(i,j,k)\in J}w_{[i, j, k]} \bar{x}_{[i,j,k]}\\ &=&\sum_{(i,j,k)\in J} |v_{[j]}||v_{[k]}| \\ &=&\sum_{i} \sum_{(j,k):(i,j,k)\in J}|v_{[j]}||v_{[k]}|\\ &= &\sum_{i} \left(\sum_{j:a_{ij}\neq 0}|v_{[j]}|\right)^2\\ &\leq& \sum_{i} \left(\sum_{j:a_{ij}\neq 0}|v_{[j]}|^2 \sum_{j:a_{ij}\neq 0}1\right)\quad\cdots(*)\\ &\leq&N\sum_{i} \sum_{j:a_{ij}\neq 0}|v_{[j]}|^2 \\ &=&N\sum_{(i,j):a_{ij}\neq 0} |v_{[j]}|^2 = N\sum_{j} |v_{[j]}|^2\sum_{i:a_{ij}\neq 0}1\\ &\leq &N^2 \sum_j |v_{[j]}|^2 = N^2||v||^2, \end{eqnarray}$$ where Cauchy-Schwarz inequality is used at $(*)$. It is a combination of changing the order of the summation and the assumption that $$ \sum_{i:a_{ij}\neq 0}1\leq N, \quad \forall j, $$ and $$ \sum_{j:a_{ij}\neq 0}1\leq N\quad\forall i. $$