Let $(x_{i,j})$ be an 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 , \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.

The bound does not even have to be explicit, namely, proving the 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.)

infinitesequence, in which case your example does indeed answer it with a rather trivial "no". But it doesn't trivialize it for the finite case. $\endgroup$6more comments