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I am studying the classical book "Ten Lectures on Wavelets" written by Ingrid Daubechies and I do not understand a specific point. I would appreciate it if someone could help me with detailed explication and new references to my study:

Let $\psi(t)$ be a square-integrable function that satisfies the zero mean condition: $\int_{\mathbb{R}} \psi(t) \, dt=0$. There is an implicit affirmation (please, see the fragment of the Daubechies book) that the zero mean condition to $\psi(t)$ implies that $\hat{\psi}(0)=0$ (here, $\hat{\psi}(t)$ is the Fourier Transform of the $\psi$ function). Why this is true? Can I demonstrate this? Besides, how are these conditions implied in Equation 2.4.1? I am a little confused about these connections.

Fragment of "Ten Lectures on Wavelets"

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    $\begingroup$ 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. $\endgroup$
    – kieransquared
    Commented Jan 12 at 22:45
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    $\begingroup$ @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. $\endgroup$
    – Yemon Choi
    Commented Jan 12 at 22:56
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    $\begingroup$ @kieransquared thanks a lot! When I read your short message I can understand. Regards! $\endgroup$
    – Luciano Magrini
    Commented Jan 12 at 23:03
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    $\begingroup$ 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 $\psi \in L^2$). $\endgroup$
    – LSpice
    Commented Jan 12 at 23:47
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    $\begingroup$ 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 $L^2(\mathbb R)$. $\endgroup$
    – Gerald Edgar
    Commented Jan 13 at 15:47

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