Let $(x_{i,j})$ be aan infinite double sequence of nonnegative real numbers, and $ 0< p<1$.
I would like to know whether one can bound from above the sum \begin{equation} \sum_{i,j} x_{i,j}^p \end{equation} in terms of \begin{equation} \sum_{i}\Big(\sum_{j} x_{i,j}\Big)^p \quad \text{and} \quad \sum_{j}\Big(\sum_{i} x_{i,j}\Big)^p \quad ? \end{equation}\begin{equation} \sum_{i}\Big(\sum_{j} x_{i,j}\Big)^p \quad , \quad \sum_{j}\Big(\sum_{i} x_{i,j}\Big)^p \quad \text{and} \quad \sum_{i,j} x_{i,j} \quad? \end{equation} The bound does not have to be tight, any upper bound will do.
My first guess was \begin{equation} \sum_{i,j} x_{i,j}^p \leq \sum_{i}\Big(\sum_{j} x_{i,j}\Big)^p + \sum_{j}\Big(\sum_{i} x_{i,j}\Big)^p + \sum_{i}\Big(\sum_{j} x_{i,j}\Big)^p \cdot \sum_{j}\Big(\sum_{i} x_{i,j}\Big)^p \end{equation} but I was unsuccessful inThe bound does not even have to be explicit, namely, proving it so farthe following statement would be useful. Let $\sum_{i,j} x_{i,j} = 1$, $\sum_{j} x_{i,j} = p_i$, and $\sum_{i} x_{i,j} = q_j$. If $\sum_{i} p_i^p < \infty$ and $\sum_{j}q_j^p < \infty$, does there exist a constant $M$ such that $\sum_{i,j} x_{i,j}^p \leq M$ for all sequences (bivariate probability distributions) $(x_{i,j})$ with marginals $(p_i)$ and $(q_j)$ ?
(This is a modified version of the problem posted a couple of days ago.)