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It is known that the Fourier transform $\mathcal{F}: L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an algebra homomorphism from $L^1$ into a sub-algebra $\mathcal(L^1(\mathbb{R}))$ of the space $C_0$ (continuous functions that vanish at $\pm\infty$).

In particular $$\mathcal{F}(f*g) = \mathcal{F}(f)\cdot\mathcal{F}(g)$$ and instead of studying convolution equations in $L^1$ we may study multiplicative equations in the Fourier image.

Can we define some homomorphism space for wavelet transform https://en.wikipedia.org/wiki/Wavelet_transform

Creator
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  • Think to both as filter banks / convolution operators. For the Fourier transform let $\hat{f}(\xi) = \int_{-\infty}^\infty f(t) e^{-2i \pi \xi t}dt$ then $f \ast e^{2i \pi \xi t}= \hat{f}(\xi) e^{2i \pi \xi t}$ and $(f \ast g)\ast e^{2i \pi \xi t}= f \ast \hat{g}(\xi) e^{2i \pi \xi t}= \hat{f}(\xi)\hat{g}(\xi) e^{2i \pi \xi t}$. For the wavelet transform replace the complex exponential by the mother wavelet. The Fourier transform diagonalizes convolution operators. – reuns May 15 '19 at 02:36
  • @reuns Thank you for the comment. May I ask what does diagonalizes convolution operators mean. Further I assume that wavelet trasform has also correspoding homomorphism. Would you please elaborate a bit as an answer. I would love to accept is as an answer. – Creator May 15 '19 at 02:46
  • @reuns I got the answer to the first question here. https://math.stackexchange.com/questions/918345/fourier-transform-as-diagonalization-of-convolution – Creator May 15 '19 at 02:52

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