Can we expect, $S(\mathbb R)\ast L^{p}(\mathbb R) \subset L^{p}(\mathbb R), (1It is well-known that $L^{1}(\mathbb R)$ is  a  closed with respect to convolution(product), that is, $L^{1}(\mathbb R)\ast L^{1}(\mathbb R)\subset L^{1}(\mathbb R),$ more specifically, if $f, g\in L^{1}(\mathbb R) $, then $f\ast g \in L^{1}(\mathbb R);$ and further more, we have $\|f\ast g\|_{L^{1}(\mathbb R)}\leq \|f\|_{L^{1}(\mathbb R)} \cdot \|g\|_{L^{1}(\mathbb R)}.$ (Infect, it is Banach algebra)
We note that, $L^{p}(\mathbb R) (1<p<\infty)$ is not closed under convolution. (It is a Banach space but not a Banach algebra with respect to convolution )
Note that, $L^{1}(\mathbb R)\ast C^{k}(\mathbb R)\subset C^{k}(\mathbb R).$ (Bit roughly speaking, this tells us that, convolution is  a smooth process)
Consider the Schwartz space, 
$\mathcal{S}(\mathbb R)= \{f\in C^{\infty}(\mathbb R): \sup_{x\in \mathbb R} (1+|x|)^{m} |\partial^{n} f(x) |< \infty, \forall m, n \in \mathbb N \} $
We also note that, $\mathcal{S(\mathbb R)}\ast \mathcal{S}(\mathbb R)\subset \mathcal{S}(\mathbb R),$ and  $\mathcal{S}(\mathbb R)\subset L^{p}(\mathbb R).$

My Question is: Can we expect, $S(\mathbb R)\ast L^{p}(\mathbb R) \subset L^{p}(\mathbb R), (1<p<\infty);$ if yes, can we expect, $\|f\ast g\|_{L^{p}(\mathbb R)} \leq \|f\|_{L^{p}(\mathbb R)} \cdot \|g\|_{L^{p}(\mathbb R)}$ for $f\in \mathcal{S}(\mathbb R), g\in L^{p}(\mathbb R), (1<p<\infty)$ ?

Thanks,
 A: The inclusion stated in the title follows from the fact that $\mathcal{S}\subset L^1$. But the inequality would involve the $L^1$ norm of $f$, not its $L^p$ norm. (Namely, Young's inequality for convolution.)  The point is, smoothness is irrelevant to $L^p$ norm estimates of this sort.
To see why you can't have $\|f\|_{L^p}$, consider $f=\chi_{[0,M]}$ (okay, this is not smooth, but you can make it taper  off smoothly at the ends  without changing the norm much). The $L^p$ norm of $f$ is $M^{1/p}$. Convolution $f*f$ is a  kind of triangle with height $M$ and base $2M$. So, its $L^p$ norm is $M^{1+1/p}$. The inequality $M^{1+1/p}\lesssim M^{1/p}M^{1/p}$ fails when $M$ is large.
A: Here's another explanation of the failure of inequality 
$$\tag{1}
\|f\ast g\|_{L^p(\mathbb{R}^N)}\le C \|f\|_{L^p(\mathbb{R}^N)}\|g\|_{L^p(\mathbb{R}^N)},\qquad f, g\in \mathcal{S}$$
for $p>1$. (The OP is about $N=1$ but there is no added difficulty in considering the general case). It is a routine application of the so-called scaling argument. 
Assume by contradiction that (1) holds for some constant $C>0$. We fix two non-vanishing functions $f$ and $g$ such that $f\ast g$ is non-vanishing (this is to avoid trivialities). Then we define
$$
f_\lambda(x)=f(\lambda x),\quad g_\lambda(x)=g(\lambda x),$$
where $\lambda >0$ is a parameter. Since $f, g\in \mathcal{S}$, their scaled versions $f_\lambda, g_\lambda$ are in $\mathcal{S}$ too. Therefore, we should have 
$$\tag{2}
\| f_\lambda \ast g_\lambda \|_{L^p}\le C \|f_\lambda\|_{L^p}\|g_\lambda\|_{L^p}.$$
However, by the change of variable formula for integrals, we also have 
$$f_\lambda \ast g_\lambda (x)=\lambda^{-N}f\ast g(\lambda x)$$
and so 
\begin{align*}
\|f_\lambda \ast g_\lambda \|_{L^p}&=\lambda^{-N\left(1+\frac{1}{p}\right)}\|f\ast g\|_{L^p}\\
\|f_\lambda \|_{L^p}&=\lambda^{-\frac{N}{p}}\|f\|_{L^p}\\
\|g_\lambda \|_{L^p}&= \lambda^{-\frac{N}{p}}\|g\|_{L^p}.
\end{align*}
Inserting those identities in (2) we obtain 
$$\tag{!!}
\lambda^{-N\left(1+\frac{1}{p}\right)} \|f\ast g\|_{L^p(\mathbb{R}^N)}\le C \lambda^{-2\frac{N}{p}}\|f\|_{L^p} \|g \|_{L^p}.$$
As you can see, the exponents on $\lambda$ are different on the left and on the right hand side unless $p=1$. This is a signal that, when $p>1$, (!!) cannot hold for all values of $\lambda>0$, and so that (2) cannot hold for all functions $f$ and $g$, no matter the value of $C$. To wit, just let $\lambda$ tend to $0$ or to $+\infty$ and see what happens. 
(Final note: In spirit, this is exactly the same answer as the one by This is much healtier. )
