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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}

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What do you call "the other inequality" and where your question originates from? – Seva Feb 18 '13 at 18:02
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} – Marcelo Feb 18 '13 at 18:10
My question originate from Lemarie's book: "recents developments in the Navier-Stokes problem" p168. – Marcelo Feb 18 '13 at 18:13
I meant $f(t)$ of course... – Yemon Choi Feb 18 '13 at 21:13
Yemon Choi, in Lemarie's book he say that \begin{align} \Vert f\Vert_{L^{\infty}}=\Vert \widehat{f}\Vert_{L^1} \end{align} – Marcelo Feb 19 '13 at 1:28
up vote 5 down vote accepted

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

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tanks so much Michael Renardy – Marcelo Feb 19 '13 at 4:37

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