Is there a double sum formula for $$\left|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\right|$$ similar to how $$\left|\sum_{n=1}^{n_1}x_n-\sum_{n=1}^{n_2}x_n\right|=\left|\sum_{n=n_1}^{n_2}x_n\right|?$$
If $m_2,n_2>m_1,n_1$, then you get basically the same formula as the single sum case. But if $m_1>m_2$ and $n_2>n_1$, or vise versa, then it does not appear to work out to a single partial double sum.
For example $$\sum_{m=1}^2\sum_{n=1}^3x_{mn}-\sum_{m=1}^3\sum_{n=1}^2x_{mn}=\sum_{m=1}^2x_{m3}-\sum_{n=1}^2x_{3n}.$$ I suppose this maybe answers my question for me, but I don't know if anyone out there has any tricks up their sleeve?
If I can show that $$\left|\sum_{m=1}^{m_1}\sum_{n=1}^{n_1}x_{mn}-\sum_{m=1}^{m_2}\sum_{n=1}^{n_2}x_{mn}\right|\leq\left|\sum\sum x_{mn}\right|$$ for some particular sum $\sum\sum x_{mn}$, then I will be on my way to proving a theorem I've been stuck on.
It might be helpful to think of $x_{m,n}$ as a function of $\mathbb{N}\times \mathbb{N}\to \mathbb{R}$. The two sums correspond to "integrating" this function on rectangles of size $m_1\times n_1$ and $m_2\times n_2$. In the case of $m_1>m_2$, $n_2>n_1$ the two rectangles are not contained within each other, so their difference will have two parts. Drawing out a picture should allow you to come up with a formula.