When the convolution is a Schwartz function?

Question

If $u$ is integrable and $v$ is in $L^p(\mathbb{R})$ ($1\leq p\leq \infty$) then $u*v\in L^p(\mathbb{R})$ (Reference: Adapted Wavelet Analysis by Mladen)

If $u$ is Schwartz function and $v$ is in $L^2(\mathbb{R})$, then $u*v$ is a Schwartz function?

Answer 1

No. In generally, we have the following result $$ L^p(\mathbb{R}^n)\ast \mathcal{S}(\mathbb{R}^n) \subseteq C^\infty(\mathbb{R}^n) \quad \forall p\in[1,+\infty] $$ See exercise 3.18 - Mitrea, D. Distribucions, Partial Differential Equations, and Harmonic Analysis.

For a counterexample: How to show convolution of an $L^p$ function and a Schwartz function is a Schwartz function