Gaussians and Young's inequality for convolutions Consider a simplified version of Young's inequality:
$$
\|f\ast g\|_p\leq \|f\|_1\|g\|_p, \quad 1< p\leq\infty
$$
$$
f\ast g\equiv \int_{\mathbb R}dy f(y)g(x-y).
$$
What strategy one should follow to determine for which functions $f\in L^1(\mathbb R)$ and $g\in L^p(\mathbb R)$ the equality holds?
I guess these should be Gaussian functions (as indicated in https://arxiv.org/abs/math/9704210), but I may be wrong.
Any hint appreciated!
 A: This inequality can be deduced in the following way. Consider
$$
(|f|\ast|g|)(x) = \int_{\mathbb{R}} |f(y)| |g(x-y)| dy. 
$$
Let's apply Hölder's inequality with respect to measure $|f(y)|dy$ to the functions $y \mapsto g(x - y)$ and $1$ with exponents $p$ and $p^\prime$ respectively. We get
$$
(|f|\ast|g|)(x) \le \left( \int_{\mathbb{R}}|g(x-y)|^p |f(y)| dy \right)^{1/p} \left( \int_{\mathbb{R}}|f(y)| dy\right)^{1/p^\prime}
$$
Then taking $L^p$ norms, we obtain
$$
\|(|f|\ast|g|)\|_{p} \le \left(\|f\|^{p-1}_1 \int_{\mathbb{R}} \int_{\mathbb{R}}  |g(x-y)|^p |f(y)| dy dx \right)^{1/p}.
$$
Using Fubini's theorem one can deduce that it's equal to
$$
\left( \|g\|_p^p \|f\|_1 \|f\|_1^{p-1}\right)^{1/p} = \|g\|_p \|f\|_1.
$$
Now we can see that in order to obtain an equality one must have an equality in Hölder's inequality. In other words, it's necessary to have
$$
|g|^{p} = c\cdot 1^{p^\prime} = 1
$$
for some constant $c$.
It cannot happen when $p < \infty$ since constant is not integrable on $\mathbb{R}$. One can see that for case $p = \infty$ desired equality holds for $g = c$.
So, equality never holds for $1 < p<\infty$ and holds with $g = c$ for $p = \infty$. 
In the article which you pointed out, authors prove stronger inequality
$$
\|f\ast g\|_p\leq C_p \|f\|_1\|g\|_p
$$
with $C_p < 1$ and apparently find functions $f,g$ for which equality takes place.
