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Let $f\in L^1{(\mathbb{R})}$. Why the Fourier Transform $\hat{f}\in L^{\infty}{(\mathbb{R})}$.

Is it because $(L^1{(\mathbb{R})})^*=L^{\infty}{(\mathbb{R})}$?

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I am ready that $(L^1{(\mathbb{R})})^*=L^{\infty}{(\mathbb{R})}$ only if If the measure $\mu$ on $\mathbb{R}$ is sigma-finite is this true? –  juaninf Oct 29 '12 at 14:13
    
This is true for every $\sigma$-finite measure, but as I pointed out in my answer, this is absolutely not related to your question. –  Ahriman Oct 29 '12 at 14:15
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1 Answer

This has nothing to do with the dual of $L^1$.

This is just an immediate consequence of the definition : for every $\xi$, $\hat{f}(\xi) = \int e^{-it \xi} f(t) \, dt$, and so $\left| \hat{f}(\xi) \right| \le \int \left|f(t) \right|dt$ which is a finite constant by definition of $L^1$.

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and ... $\hat{f}(\xi) \in L^1$? –  juaninf Oct 29 '12 at 15:54
    
@Juan Does $\hat{f} \in L^1$ when $f$ is the characteristic function of $[-1,1]$ ? –  Ahriman Oct 29 '12 at 16:02
    
not necessarily –  juaninf Oct 29 '12 at 16:04
    
sorry the in the first comment the question is $\hat{f} \in L^1$? –  juaninf Oct 29 '12 at 16:05
    
And so my comment gives a counterexample showing that $\hat{f}$ doesn't belong necessarily to $L^1$ if $f \in L^1$. –  Ahriman Oct 29 '12 at 16:10
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