Double Sum Identity Problem Solution

[CompuChip]

Hi. The problem is as follows:

Homework Statement

Let m and n be integers, we may assume that (if they are not equal), m is the smallest. Then
\sum_{i=0}^m \sum _{j=0}^n f((m+n)-2 (i+j)) = \sum_{i=0}^m \sum _{j=0}^{-2 i+m+n} f((m+n)-2 (i+j))
for some sequence f(k)_k.

Homework Equations

Anything you can think of, but probably just the standard manipulations on sums and some smart rewriting and/or separating different cases.

The Attempt at a Solution

I have reduced from a given question to the above statement, and I'm quite positive that I haven't made a mistake in this -- so the given statement should be true. It doesn't really look true to me though, so I plugged this into Mathematica: it seems to hold for any two m \le n I choose, but I just don't see how to prove this. I think it involves a manipulation with the double sum, because if I remove the outer sum and consider i a fixed number in the rest of the expression, it doesn't seem to hold anymore (I always need one or more of the f(n) to be zero, which they aren't in general).
So I think the given information now is necessary and sufficient (e.g. no extra information on f(n) or m and n should be needed).

 

[EnumaElish]

I'd start with concrete examples (m = 1, n = 2, etc.) to see how exactly it is working.

 

[CompuChip]

I already did, and I also explicitly worked out the multiplicities of the various f(m + n - 2 i) of the left hand side, I found it to be equal to
<br /> \sum_{i = 0}^{m - 1} (i + 1) f(m + n - 2i) +<br /> {\sum_{k = m - n}^{n - m}}&#039; (m + 1) f(k) +<br /> \sum_{i = 0}^{m - 1} (i + 1) f(-(m + n - 2i)),<br />
where the prime in the second sum denotes summation over odd values only (so m - n, m - n + 2, ...).So I could just do the same for the right hand side and show that the multiplicity of each term is equal, but I was sort of hoping that the result can also be obtained by just manipulating the sums (e.g. operations like shifting indices, switching summation orders, etc.).