Hint about a identity Who has hint how to prove:
$\sum_{n=0}^N \sum_{k=1}^N g_k y_n = \sum_{n=1}^N \sum_{k=1}^n g_k y_{n-k}+\sum_{n=1}^N\sum_{k=1}^n g_n y_{N-k+1}$
Thank in advance!

It's almost always helpful to expand both sides to see where all the terms come from:  
LHS: We can factor out $g_{k}$ from the inner sum, since it is independent of both $n$ and $N$. Thus, we have $g_{1}(y_{0}+y_{1}+\ldots+y_{N})+g_{2}(y_{0}+\ldots+y_{N})+\ldots+g_{N}(y_{0}+\ldots +y_{N})$, which we can easily see is $$y_{0}(g_{1}+g_{2}+\ldots+g_{N})+y_{1}(g_{1}+\ldots+g_{N})+\ldots+y_{N}(g_{1}+\ldots+g_{N})$$   
RHS: I think your first sum is incorrect; it should read $\sum_{n=1}^{N} \sum_{k=1}^{n} g_{k}y_{n-k}$. If we expand this first sum, where each set of square brackets contains the next value for $n$ (the first set for $n=1$, the next for $n=2$, etc.), we get the following:$[g_{1}y_{0}]+[g_{1}y_{1}+g_{2}y_{0}]+[g_{1}y_{2}+g_{2}y_{1}+g_{3}y_{0}]+\ldots+[g_{1}y_{N-1}+g_{2}y_{N-2}+\ldots g_{N}y_{0}]$
Which, after rearranging, is equal to $y_{0}(g_{1}+g_{2}+\ldots+g_{N})+y_{1}(g_{1}+g_{2}+\ldots+g_{N-1})+\ldots y_{N-1}g_{1}$
Before expanding the second sum, let's see what we have. Each of the $y_{k}$ Has $N-k$ of the terms we want it to be multiplied by. So $y_{0}$ is being multiplied by all the $g_{k}$ we require, $y_{1}$ by all but $g_{N}$, and so on. Let us move onto the second sum.
Expanding similarly, $[g_{1}y_{N}]+[g_{2}y_{N}+g_{2}y_{N-1}]+\ldots+[g_{N}y_{N}+g_{N}y_{N-1}+\ldots g_{N}y_{1}]$. By rearranging, we have $y_{1}g_{N}+y_{2}(g_{N-1}+g_{N})+\ldots+y_{N}(g_{1}+g_{2}+\ldots+g_{N})$  
Add the two sums on the RHS and collect your terms, and your identity follows.
