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I'm trying to understand why the wavefront set of the distribution $1/(x+i0)$ is given by $\left\{(x,\xi):\xi=0\ \textrm{or}\ x=0,\xi\geq0\right\}$, and why that of $1/(x-i0)$ is $\left\{(x,\xi):\xi=0\ \textrm{or}\ x=0,\xi\leq0\right\}$.

Also, I'm trying to figure out the wavefront set of the principal value of $1/x$.

Can anyone please help?

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The Fourier transform of $1/(x+i\varepsilon)\;$, $\varepsilon>0$ is equal to $$ -i \sqrt{2 \pi } \theta (-\xi ) e^{\xi \epsilon }. $$ Taking the limit as $\varepsilon\to0+$ one gets the Fourier transform of $1/(x+i0)$: $$ g_+(x)=-i \sqrt{2 \pi } \theta (-\xi ). $$ Now by the definition of the wave front set we have to consider the rate of decrease at infinity of $$ F[g_+\varphi]=F[g_+]*F[\varphi],\ $$ where $\varphi\in \mathcal D$, the support of $\varphi$ is in some neighbourhood of the origin, $\varphi(0)\ne0$. Since $F[\varphi]$ decreases at infinity faster than any polynomial, the convolution $F[g_+]*F[\varphi](\xi )$ has the same property as $\xi \to+\infty$. But not when $\xi \to-\infty$ because $$\lim_{\xi \to-\infty}F[g_+]*F[\varphi](\xi )=-i \sqrt{2 \pi } \varphi(0)\ne0.$$ The assertion for $1/(x-i0)$ is checked in the same way. As for the principal value of $1/x$, in one dimensional case there are only two directions. And both have to belong to the WF set due to the symmetry (oddness) of the distribution. The only other possibility is that both directions don't belong to WF set. But it would mean that the origin does not belong to the singular support of $1/x$, a contradiction.

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What is meant by $\theta(-\xi)$? – Dave Oct 12 '11 at 17:43
@Dave $\theta$ is the Heaviside step function – Andrew Oct 12 '11 at 18:05

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