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!

We have 
$$
\begin{align}
\sum_{k=1}^N \sum_{n^\prime=0}^N g_k y_n
& = \sum_{\substack{1\leq n\leq N\\ 1\leq k \leq N}} g_k y_n 
\\
& = \sum_{(k,n)\in G} g_k y_n 
\\
\end{align}
$$
Here $G=\{ (k,n) \in\mathbb{N}\times\mathbb{N} : 1\leq n\leq N, 1\leq k \leq N\}$. Set 
$$
\begin{align}
G_1 &=\{ (k,n) \in\mathbb{N}\times\mathbb{N} : 1\leq n\leq k, 1\leq k \leq N, \} \\
     &   \\
G_2 &= \{ (k,n) \in\mathbb{N}\times\mathbb{N} : k+1\leq n\leq N, 1\leq k \leq N\}.\\
\end{align}
$$
Then $G=G_1\cup G_2$, $G_1\cap G_2=\emptyset$ and
$$ 
\begin{align}
\sum_{k=1}^N \sum_{n=0}^N g_k y_n 
& = \sum_{(k,n)\in G} g_k y_n 
\\
& = \sum_{(k,n)\in G_1} g_k y_n + \sum_{(k,n)\in G_2} g_k y_n 
\\
& = \sum_{\substack{1\leq n\leq k\\ 1\leq k \leq N}} g_k y_n 
+
\sum_{\substack{k+1\leq n\leq N\\ 1\leq k \leq N}} g_k y_n 
\\
& = \sum_{1\leq k \leq N}\;\;\sum_{1\leq n\leq k} g_k y_n 
+
\sum_{1\leq k \leq N}\;\;\sum_{k+1\leq n\leq N} g_k y_n 
\\
& = \sum_{1\leq k \leq N}\;\;\sum_{1\leq n\leq k} g_k y_n 
+
\sum_{1\leq k \leq N}\;\;\sum_{1\leq n-k\leq N-k} g_k y_n 
\\
\end{align}
$$
Making the "change of variables" $n=n^\prime+k$:
$$
\begin{align}
\sum_{k=1}^N \sum_{n=0}^N g_k y_n 
& = \sum_{1\leq k \leq N}\;\;\sum_{1\leq n\leq k} g_k y_n 
+
\sum_{1\leq k \leq N}\;\;\sum_{1\leq n^\prime \leq N-k} g_{k^\prime} y_{n^\prime+k} 
\\
\end{align}
$$
The change in notation $n=n^\prime$ does not interfere in the value of the sum we now
$$
\begin{align}
\sum_{k=1}^N \sum_{n^\prime=0}^N g_k y_n 
& = \sum_{1\leq k \leq N}\;\;\sum_{1\leq n^\prime\leq k} g_k y_{n^\prime} 
+
\sum_{1\leq k \leq N}\;\;\sum_{1\leq n^\prime \leq N-k} g_{k^\prime} y_{n^\prime+k} 
\\
\end{align}
$$
