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Holy moly I spelled Titchmarsh's name wrong.
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Willie Wong
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Yes, there is one such example: $u \equiv 0$.


The answer above is not facetious! That $u$ is in fact the only example (modulo measure zero modifications).

By Titchmarsch'sTitchmarsh's theorem, if $u\in L^2(\mathbb{R})$ and its Fourier support is on the positive real line, $u$ must be equal to the trace of some holomorphic function $F$ defined on the upper half plane.

If $u$ itself further vanishes on the left half line, which has positive measure, by the Luzin-Privalov Theorem the function $F$ must vanish identically. Hence the only function satisfying your condition is identically zero.

Yes, there is one such example: $u \equiv 0$.


The answer above is not facetious! That $u$ is in fact the only example (modulo measure zero modifications).

By Titchmarsch's theorem, if $u\in L^2(\mathbb{R})$ and its Fourier support is on the positive real line, $u$ must be equal to the trace of some holomorphic function $F$ defined on the upper half plane.

If $u$ itself further vanishes on the left half line, which has positive measure, by the Luzin-Privalov Theorem the function $F$ must vanish identically. Hence the only function satisfying your condition is identically zero.

Yes, there is one such example: $u \equiv 0$.


The answer above is not facetious! That $u$ is in fact the only example (modulo measure zero modifications).

By Titchmarsh's theorem, if $u\in L^2(\mathbb{R})$ and its Fourier support is on the positive real line, $u$ must be equal to the trace of some holomorphic function $F$ defined on the upper half plane.

If $u$ itself further vanishes on the left half line, which has positive measure, by the Luzin-Privalov Theorem the function $F$ must vanish identically. Hence the only function satisfying your condition is identically zero.

Source Link
Willie Wong
  • 39.1k
  • 4
  • 94
  • 176

Yes, there is one such example: $u \equiv 0$.


The answer above is not facetious! That $u$ is in fact the only example (modulo measure zero modifications).

By Titchmarsch's theorem, if $u\in L^2(\mathbb{R})$ and its Fourier support is on the positive real line, $u$ must be equal to the trace of some holomorphic function $F$ defined on the upper half plane.

If $u$ itself further vanishes on the left half line, which has positive measure, by the Luzin-Privalov Theorem the function $F$ must vanish identically. Hence the only function satisfying your condition is identically zero.