1
$\begingroup$

Let $A:=\{f\in C^1(\mathbb{R}): \hat{f}, \hat{f'} \in L^1(\mathbb{R})\}$, where $\hat{f}$ is the Fourier transform of $f$. Then is it true that Schwartz space $\mathcal{S}(\mathbb{R})$ is dense in $A$ w.r.t. $\lVert f\rVert:= \lVert\hat{f}\rVert_1+\lVert\hat{f'}\rVert_1$?

I only know that $\mathcal{S}(\mathbb{R})$ is dense in $L^1(\mathbb{R})$.

So, can you please guide me on this problem?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ This is equivalent to ask if the Schwartz space is dense in $L^1(1+|\xi|) d\xi$ and it is true. Take a functions, approximate first with one with compact support and then use convolution: when the support is fixed you do not see the weight. $\endgroup$
    – Giorgio Metafune
    Commented May 19 at 16:29
  • $\begingroup$ It is unclear why my question is equivalent to the Schwartz space being dense in $L^1(1+|ξ|)dξ$. Can you please give a bit of details? $\endgroup$
    – mathlover
    Commented May 19 at 17:41
  • 1
    $\begingroup$ Your have $\|f\|=\|\hat f\|_1+\|\xi \hat f\|_1=\|(1+|\xi|)\hat f\|_1$. If you approximate $\hat f$ in the weighted $L^1$ norm above, by Schwartz functions $g_n$, then take $f_n$ the inverse transforms of $g_n$. $\endgroup$
    – Giorgio Metafune
    Commented May 19 at 17:48
  • 1
    $\begingroup$ The definition of $A$ is not very clear. If $f$ is not of temperate growth I don't know how to define $\hat{f}$. $\endgroup$
    – an_ordinary_mathematician
    Commented May 19 at 18:38
  • $\begingroup$ We implicitly assume all necessary conditions that are required to exist $\hat{f}$ and $\hat{f'}$. $\endgroup$
    – mathlover
    Commented May 20 at 11:39

0

Reset to default