Let $L^2(\mathbb{T}^3)$ be the L^2 space on $3$-dimensional torus. That is, I am considering periodic functions.
Then, it is well-known that the there exists a smooth orthonormal basis of $L^2(\mathbb{T}^3)$ consisting of eigenfunctions of the Laplacian $-\Delta$.
Let us denote such an eigenbasis by $\{f_n\}$ with corresponding eigenvalues $\lambda_n$.
Now, let us suppose that $\lambda_1 \neq \lambda_2$ and consider the convolution $f_1 * f_2$, which is clearly smooth and periodic on $\mathbb{T}^3$.
But my observation is that, \begin{equation} -\Delta (f_1 * f_2)=(-\Delta f_1) * f_2 =f_1 *(-\Delta f_2) \end{equation} so that, $(\lambda_1 - \lambda_2)f_1 * f_2 =0$.
However, I already assumed that $\lambda_1 \neq \lambda_2$. Thus, $f_1 * f_2=0$.
Is my reasoning correct? That is, convolution of eigenvectors for $-\Delta$ corresponding to different eigenvalues is just identically zero?