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If we define the Heaviside function in the standard way $H(x)=\begin{cases} \ 1 & x\geq 0 \\ \ 0& x<0 \end{cases}$

Then I want to find the Wavefront set where I am using the definition as: The wavefront set of a distribution is the set of points $(x,k)\in \mathbb{R}^n \times (\mathbb{R}^n - \{ 0 \})$ which are not regular directed.

Where by regular directed i mean: For a distribution $u\in \mathcal{D}'(\mathbb{R}^n)$ a point $(x,k)\in \mathbb{R}^n \times \mathbb{R}^n - \{0\} $ is called $\text{a regular directed point}$ of u if and only if $\exists$ a function $f\in \mathcal{D}(\mathbb{R}^n)$ with $f(x)=1$ and $\exists$ a closed conical neighborhood $V\subset \mathbb{R}^n$ of k, such that $\widehat{fu}$ is fast decreasing on V.

Minkowski 2.0
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You use that $WF(Pu) \subset WF(u)$ for every differential operator with smooth coefficients $P$ (you can find the proof in almost any text concerning wavefronts e.g. The Analysis of Linear Partial Differential Operators I by Hormander). Then note that $H$ is smooth everywhere except in the origin, so its singular support is just $\lbrace 0 \rbrace$. Finally $\partial H = \delta_0$, thus $WF(\delta_0) \subset WF(H)$ and since $WF(\delta_0)= \lbrace 0 \rbrace \times \mathbb{R} \backslash \lbrace 0 \rbrace$ you get that $WF(H)=\lbrace 0 \rbrace \times \mathbb{R} \backslash \lbrace 0 \rbrace$.

Baol
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  • This is close, but not quite complete: one still needs to show the wavefront set of $\delta_0$. Ultimately, $|\exp(i\xi)| \in \Theta(1)$. The property that $\text{WF}(\partial u) \subset \text{WF}(u)$ (understood distributionally) is rather straightforward under Fourier transform. Alternatively, one can use directly the f.t. of H to see that it is $\Theta(1/|\xi|)$; the point is, a decay rate comparison must come in somewhere to conclude this. – akkapi May 09 '21 at 23:53