1
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ What do you call "the other inequality" and where your question originates from? $\endgroup$ – Seva Feb 18 '13 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} $\endgroup$ – Marcelo Feb 18 '13 at 18:10
  • $\begingroup$ My question originate from Lemarie's book: "recents developments in the Navier-Stokes problem" p168. $\endgroup$ – Marcelo Feb 18 '13 at 18:13
  • $\begingroup$ I meant $f(t)$ of course... $\endgroup$ – Yemon Choi Feb 18 '13 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} $\endgroup$ – Marcelo Feb 19 '13 at 1:28
9
$\begingroup$

If $\hat f$ is nonnegative, then (up to a factor), $$f(0)=\int \hat f=\Vert \hat f \Vert_1 = \Vert f \Vert_\infty.$$

$\endgroup$
  • $\begingroup$ tanks so much Michael Renardy $\endgroup$ – Marcelo Feb 19 '13 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.