Research proof index

Mellin convolution is commutative at positive arguments. For functions $f,g:\mathbb R\to\mathbb R$ or $\mathbb C$ and every $x>0$, $(f*_M g)(x)=(g*_M f)(x),$ where $(f*_M g)(x)=\int_0^\infty f(y)g(x/y)\,dy/y$.

Let $\nu$ be continuously differentiable and supported in $[1/2,2]$. For $\varepsilon>0$ and $\operatorname{Re}s>0$, the Mellin transform of the project's smoothed cutoff satisfies $\mathcal M(\widetilde{1_\varepsilon})(s)=\frac{1}{s}\,\mathcal M(\nu)(\varepsilon s).$