0
$\begingroup$

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

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

  • 0
    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
  • 0
    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

1 Answers 1