# Equality in Young's inequality for convolution

I am trying to understand how sharp Young's inequality for convolution is. The inequality says $||f \ast g||_r \leq ||f||_p ||g||_q$ where as $1/p+1/q = 1+1/r$.

Actually, there are a couple of papers (for example: Sharpness in Young's inequality for convolution) talking about the case of $p, q>1$, and the answer is that we can find a constant $C<1$ such that $||f \ast g||_r \leq C||f||_p ||g||_q$

However, I believe that when $f\in L^1$ and nonnegative, we will have a sharp $||f \ast g||_p \leq ||f||_1 ||g||_p$. Does anyone know a proof of this?

To be more precise, can we construct a sequence $\{g_n\}$ of $L^p$ functions, such that $||f\ast g_n||_p/||g_n||_p\rightarrow ||f||_1$ as $n\rightarrow \infty$?

The sharp Young's inequality reads: $$\|f \ast g \|_r \leq \left( A_p A_q A_{r'} \right)^n \|f\|_p\|g\|_q$$ where $A_s = \left( \frac{s^{1/s}}{s'^{1/s'}} \right)$, $\frac{1}{s} + \frac{1}{s'} = 1$, $f$ and $g$ are functions on $n$ dimensions, and $1 \leq p,\,q,\,r\leq \infty$, as usual. In order to show that your inequality is sharp, it's only necessary to note that $A_s A_{s'}=A_1=1$.
This inequality is saturated whenever $f$ and $g$ are both Gaussians, according to the paper.