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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?2012-10-29
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    This is true for every $\sigma$-finite measure, but as I pointed out in my answer, this is absolutely not related to your question.2012-10-29

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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 ... $\hat{f}(\xi) \in L^1$?2012-10-29
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    @Juan Does $\hat{f} \in L^1$ when $f$ is the characteristic function of $[-1,1]$ ?2012-10-29
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    not necessarily2012-10-29
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    sorry the in the first comment the question is $\hat{f} \in L^1$?2012-10-29
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    And so my comment gives a counterexample showing that $\hat{f}$ doesn't belong necessarily to $L^1$ if $f \in L^1$.2012-10-29
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    $\int|\hat{f}(\xi)| d\xi = \int|\int e^{-it \xi} f(t) \, dt|d\xi \leq \int\int | f(t) |\, dtd\xi$, This $\int\int | f(t) |\, dtd\xi$ expression is infinite?2012-10-29
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    What mean characteristic function of [-1,1]?2012-10-29