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1 vote

Fourier transform of a function of bounded variation

If $f'$ is of bounded variation, then $\hat{f}$ will be integrable. To see this, note that by your assumptions, $f''$ (in the sense of tempered distributions) is a finite complex-valued measure, so th …
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2 votes
0 answers
802 views

Existence of unbounded $M \subset \Bbb{R}$ of finite measure s.t. $1_M$ is $L^p$-Fourier mul...

I would like to know if there is a measurable set $M \subset \Bbb{R}$ such that $M$ has finite Lebesgue measure $0 < \lambda(M) < \infty$, $M$ is unbounded in the sense that $\lambda(M \setminus [-r …
5 votes
0 answers
209 views

Existence of $A\subset\Bbb{R}^n$ of finite measure and $\hat{1_A}\in\bigcap_{q>1}L^q$, but s...

I am interested in the following somewhat obscure question: Is there some $n \in \Bbb{N}$, and a set $A \subset \Bbb{R}^n$ of finite measure such that the Fourier transform $\widehat{1_A}$ of its …
6 votes
2 answers
2k views

Reverse Hausdorff Young for nonnegative functions

The classical Hausdorff-Young inequality states that $$ \Vert \widehat{f} \Vert_{p'} \leq \Vert f \Vert_p \text{ for } 1 \leq p \leq 2. $$ For $p=2$, we even have equality due to Plancherel. If we …