Let $\widehat{f}(\xi)$ be Fourier transform of $f$ given by \begin{align} \widehat{f}(\xi)=\int_{\mathbb{R}^n} e^{-ix\cdot\xi}f(x)dx. \end{align} Suppose that $\widehat{f}(\xi)$ is nonnegative and locally integrable function, easily seems (by inverse Fourier transform) that \begin{align} \Vert f\Vert_{L^{\infty}} \leq \Vert \widehat{f}\Vert_{L^1}. \end{align} How to show that there is a positive constant $c>0$ such that \begin{align} \Vert \widehat{f}\Vert_{L^1}\leq c \Vert f\Vert_{L^{\infty}}. \end{align}
$\begingroup$
$\endgroup$
6
-
$\begingroup$ What do you call "the other inequality" and where your question originates from? $\endgroup$– SevaFeb 18, 2013 at 18:02
-
$\begingroup$ How to show that there is a positive constant $c>0$ such that \begin{align} \Vert \widehat{f}\Vert_{L^1}\leq c \Vert f\Vert_{L^{\infty}} \end{align} $\endgroup$– user9663Feb 18, 2013 at 18:10
-
$\begingroup$ My question originate from Lemarie's book: "recents developments in the Navier-Stokes problem" p168. $\endgroup$– user9663Feb 18, 2013 at 18:13
-
$\begingroup$ I meant $f(t)$ of course... $\endgroup$– Yemon ChoiFeb 18, 2013 at 21:13
-
$\begingroup$ Yemon Choi, in Lemarie's book he say that \begin{align} \Vert f\Vert_{L^{\infty}}=\Vert \widehat{f}\Vert_{L^1} \end{align} $\endgroup$– user9663Feb 19, 2013 at 1:28
|
Show 1 more comment
