Sum of the generalized convolution product of two signals

Question

I am trying to prove this property, which states that the sum of the generalized convolution of two product, $$\sum_{n=1}^N (f*g)(n)=\sqrt{N}\hat{f}(0)\hat{g}(0)$$ The property is found on the article "Vertex-frequency analysis on graphs". I tried proving by definition, where $$\sum_{n=1}^N (f*g)(n)=\sum_{n=1}^N \sum_{\ell=0}^{N-1}\hat{f}(\lambda_\ell)\hat{g}(\lambda_\ell)\chi_\ell(n)$$. I also converted the inner summation by getting the fourier transform of it, so I have now $$\sum_{n=1}^N (f*g)(n)=\sum_{n=1}^N f(n)g(n)$$ After this, I don't know how to proceed. I just know that $\sqrt{N}$ has something to do with the eigenvalues. Any ideas?

Answer 1

I get it now, I realized the process I did above was wrong. Here's my answer to the problem. $$\sum_{n=1}^N (f*g)(n)=\sum_{n=1}^N\sum_{\ell=0}^{N-1}\hat{f}(\lambda_\ell)\hat{g}(\lambda_\ell)\chi_\ell (n)$$ After that we multiply $\frac{1}{\sqrt{N}}$ to both sides. We now have $$\frac{1}{\sqrt{N}}\sum_{n=1}^N (f*g)(n)=\frac{1}{\sqrt{N}}\sum_{n=1}^N\sum_{\ell=0}^{N-1}\hat{f}(\lambda_\ell)\hat{g}(\lambda_\ell)\chi_\ell (n)$$ $$=\frac{1}{\sqrt{N}}\sum_{\ell=0}^N\sum_{n=1}^{N-1}\hat{f}(\lambda_\ell)\hat{g}(\lambda_\ell)\chi_\ell (n)$$ $$=\frac{1}{\sqrt{N}}\sum_{\ell=1}^{N-1}\hat{f}(\lambda_\ell)\hat{g}(\lambda_\ell)\sum_{n=1}^{N}\chi_\ell (n)$$ Since given from the article that $\chi_0=\frac{1}{\sqrt{N}}$, we have now, $$=\sum_{\ell=0}^{N-1}\hat{f}(\lambda_\ell)\hat{g}(\lambda_\ell)\sum_{n=1}^{N}\chi_0\chi_\ell (n)$$ The inner summation defines a orthonormal basis, that is, the dot product is equal to 0 if $0\neq\ell$ otherwise 1; and given than $\lambda_0=0$. Hence we have, $$\frac{1}{\sqrt{N}}\sum_{n=1}^N (f*g)(n)=\hat{f}(0)\hat{g}(0)$$. Finally, simplifying we get the equation: $$\sum_{n=1}^N (f*g)(n)=\sqrt{N}\hat{f}(0)\hat{g}(0)$$