Timeline for Why for $\psi$ square-integrable function the zero mean condition is equivalent to $\hat{\psi}(0) = 0$?
Current License: CC BY-SA 4.0
12 events
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| Jan 13 at 15:47 | comment | added | Gerald Edgar | A nonsense(?) answer.... Plancherel's theorem yields $$\int \psi(t)\;dt = \langle\psi,1\rangle= \langle\widehat{\psi},\delta\rangle = \widehat{\psi}(0),$$ since $\hat{1} = \delta$. But Plancherel is for $L^2$ functions, and neither $1$ nor $\delta$ is in $L^2(\mathbb R)$. | |
| Jan 12 at 23:48 | history | edited | LSpice | CC BY-SA 4.0 |
Missing right quote
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| Jan 12 at 23:47 | comment | added | LSpice | The claim in the excerpted image asserts the equivalence $\hat\psi(0) = 0 \iff \int \psi(x)\mathrm dx = 0$ only for $\psi \in L^1$ (although the surrounding context suggests that we also mean to assume that $\psi \in L^2$). | |
| Jan 12 at 23:13 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 3 characters in body; edited title
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| Jan 12 at 23:11 | review | Close votes | |||
| Jan 22 at 3:02 | |||||
| Jan 12 at 23:07 | comment | added | kieransquared | @YemonChoi You’re absolutely right. $\psi\in L^1\cap L^2$ should be enough. | |
| Jan 12 at 23:05 | comment | added | Luciano Magrini | @YemonChoi thanks. I will search for concepts of regularity and approximation arguments to understand how Equation 2.4.1 can be derived from the mean zero condition. | |
| Jan 12 at 23:03 | comment | added | Luciano Magrini | @kieransquared thanks a lot! When I read your short message I can understand. Regards! | |
| Jan 12 at 22:56 | comment | added | Yemon Choi | @kieransquared I think there has to be some tacit regularity or integrability assumption already on $\psi$. Point evaluations are not well-defined on elements of $L^2(\widehat{\mathbb R})$, so to even speak of $\widehat{\psi}(0)$ some extra assumption/conditions are needed. | |
| Jan 12 at 22:45 | comment | added | kieransquared | For an $L^1$ function this is clear from the definition of the Fourier transform: $\hat{\psi}(0) = \int e^{-ix\cdot 0} \psi(x)dx = 0$. For an $L^2$ function you can extend this fact by a straightforward approximation argument. | |
| S Jan 12 at 22:23 | review | First questions | |||
| Jan 12 at 22:55 | |||||
| S Jan 12 at 22:23 | history | asked | Luciano Magrini | CC BY-SA 4.0 |
