Can we expect, $S(\mathbb R)\ast L^{p}(\mathbb R) \subset L^{p}(\mathbb R), (1<p<\infty) $?

Question

It 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,

Answer 1

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. )

Answer 2

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.