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fourier analysis - Convolution theorem with distributions - Mathematics Stack Exchange

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Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Let 𝜑: ℝ 𝑛 →[0,∞) 𝜑 : 𝑅 𝑛 → [ 0 , ∞ ) be compactly supported and 𝐶 ∞ 𝐶 ∞ . Define 𝑘 𝑠 (𝑥)=|𝑥 | −𝛼 𝑘 𝑠 ( 𝑥 ) = | 𝑥 | − 𝛼 for 𝑥∈ ℝ 𝑛 𝑥 ∈ 𝑅 𝑛 , where 0<𝛼<𝑛 0 < 𝛼 < 𝑛 . I know that, as tempered distributions, ( 𝑘 𝛼 )= 𝑘 𝑛−𝛼 𝐹 ( 𝑘 𝛼 ) = 𝑘 𝑛 − 𝛼 and (𝜑∗ 𝑘 𝛼 )=(𝜑)( 𝑘 𝛼 ). 𝐹 ( 𝜑 ∗ 𝑘 𝛼 ) = 𝐹 ( 𝜑 ) 𝐹 ( 𝑘 𝛼 ) . Here  𝐹 denotes the Fourier transform. Question Is the following true? If yes, how can it be proved? ∫ ℝ 𝑛 |(𝜑∗ 𝑘 𝛼 ) | 2 𝑑𝑥= ∫ ℝ 𝑛 |(𝜑)( 𝑘 𝛼

fourier analysis - Convolution theorem with distributions - Mathematics Stack Exchange Stack Internal Knowledge at work Bring the best of human thought and AI automation together at your work. Explore Stack Internal Convolution theorem with distributions Ask Question Asked 8 years, 3 months ago Modified 8 years, 3 months ago Viewed 2k times 4 $\begingroup$ Let $\varphi:\mathbb{R}^n \to [0,\infty)$ be compactly supported and $C^{\infty}$. Define $k_s(x) = |x|^{-\alpha}$ for $x \in \mathbb{R}^n$, where $0 < \alpha < n$. I know that, as tempered distributions, $$ \mathcal{F}(k_\alpha) = k_{n-\alp

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