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Mark Lewko
Mark Lewko
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The only result along these lines that I am aware of is due to Vebedev and Olevskii (Idempotents of Fourier multiplier algebra. Geom. Funct. Anal. 4 (1994), no. 5, 539--544. ). Their result states that if is a set E \subset R^d does not agree (almost everywhere) with an open set, then it can not be a L^p multiplier for p \neq 2.

The only result along these lines that I am aware of is due to Vebedev and Olevskii (Idempotents of Fourier multiplier algebra. Geom. Funct. Anal. 4 (1994), no. 5, 539--544. ). Their result states that if is a set E \subset R^d does not agree (almost everywhere) with an open set, then it can not be a L^p multiplier for p \neq 2.

The only result along these lines that I am aware of is due to Vebedev and Olevskii (Idempotents of Fourier multiplier algebra. Geom. Funct. Anal. 4 (1994), no. 5, 539--544. ). Their result states that if a set E \subset R^d does not agree (almost everywhere) with an open set, then it can not be a L^p multiplier for p \neq 2.

Source Link
Mark Lewko
Mark Lewko
  • 13k
  • 1
  • 55
  • 87

The only result along these lines that I am aware of is due to Vebedev and Olevskii (Idempotents of Fourier multiplier algebra. Geom. Funct. Anal. 4 (1994), no. 5, 539--544. ). Their result states that if is a set E \subset R^d does not agree (almost everywhere) with an open set, then it can not be a L^p multiplier for p \neq 2.