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I have a relation of the form $$\omega = f\ast \mu \qquad (1)$$ where $\omega $ and $\mu $ are distributions in $\mathcal {D}'(\mathbb{R})$ and $$f(x) = H(x-1)(x^2 - 1)^{1/2}x$$ with $H(x) = 1_{\mathbb{R}_+}$ denoting the Heaviside function.

My goal is to show that $\mathrm{WF}_a (\omega ) = \mathrm{WF}_a (\mu )$ where $\mathrm{WF}_a$ is the analytical wave front set, and for this I would like to invert (1), i.e. solve it for $\mu $. By taking Fourier transforms I get $$ \mu = C \widehat {\Big (\frac{1}{\hat {f}}\Big)} \ast \omega $$ for some constant C. Here I would thus like to conclude that if $\omega $ extends to a holomorphic function near some $x\in \mathbb{R}$ then the same goes for $\mu $.

My problem is I have no clue what the transform of $1/\hat{f}$ is.

EDIT: I have changed from $H(x)$ to $H(x-1)$ in the definition of $f$.

edited Mar 28 2012 at 9:24

asked Mar 27 2012 at 9:10

Alex A
274●2●8

According to Maple $\hat{f}=C\frac{K_1(i\xi )}{\xi }$ where $K_1$ is a modified Bessel function. I think $K_1(i\xi ) = C_0H_1^{(1)}(-\xi )$ where $H_1^{(1)}$ is a Hankel function. Thus I'm interested in analytic extensions of the transform of $\frac{\xi }{H_1^{(1)}(-\xi )}$. – Alex A Mar 28 2012 at 9:58

1

The question posed is to determine the Fourier transform of a specific distribution and not inverting convolution as your title suggests. BTW, you don't need to invert the convolution to show equality of WF sets. Since the dominant term of the $\hat{f}$ is a constant and it is multiplying the fourier transform of $\mu$, we should be able to say that the WFs are same, irrespective of we know $\mu$ or not. – Uday Mar 28 2012 at 10:52

Inverting convolution

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