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LSpice
LSpice
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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 be $\psi(t)$ 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 WaveletsWavelets"

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 be $\psi(t)$ 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

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"

added 3 characters in body; edited title
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Michael Hardy
Michael Hardy
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Why for $\psi$ square-integrable function the zero mean condition is equivalent to $\hat{\psi}(0) = 00$?

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 be $\psi(t)$ a square-integrable function that satisfies the zero mean condition: $\int_{\mathbb{R}} \psi(t) dt=0$$\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

Why for $\psi$ square-integrable function the zero mean condition is equivalent to $\hat{\psi}(0) = 0?

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 be $\psi(t)$ 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

Why for $\psi$ square-integrable function the zero mean condition is equivalent to $\hat{\psi}(0) = 0$?

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 be $\psi(t)$ 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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Luciano Magrini
Luciano Magrini
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Why for $\psi$ square-integrable function the zero mean condition is equivalent to $\hat{\psi}(0) = 0?

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 be $\psi(t)$ 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