It is a standard exercise to show that if $X\subseteq\mathbb{R}^n$ has smooth boundary, then the characteristic function $1_X$ has wavefront set $$\{(x,\xi)\in\partial X\times\mathbb{R}^n\setminus\{0\}:\xi\text{ is normal to }\partial X\}.$$ This is shown by locally "flattening" to reduce to the case of the upper half space.

My question is what happens when $X$ does not have smooth boundary, such as the interior of a fractal curve. For example, if $X$ is the interior of the Koch snowflake, how would one compute the wavefront set of $1_X$?


1 Answer 1


Let me first begin with an elementary example, taking $X=[0,1]^2$ in $\mathbb R^2$. It is then easy to see directly that the wave-front-set of $\mathbb 1_X$ is everywhere the conormal bundle except at the four corners $c_j$: $$ WF\mathbb 1_X=\cup_{1\le j\le 4}(c_j;\mathbb R^2\backslash {(0,0)})\cup\text{conormal bundle at the smooth points}. $$ To prove this, it is enough to check the wave-front-set of $H(x_1)H(x_2)$ with $H=\mathbb 1_{[0,+\infty)}$ which amounts to check the wave-front-set of $\delta_0(x_1)\otimes\delta_0(x_2)$ which contains at $(0,0)$ every direction since the Fourier transform is 1.

There are of course (much) more refined results. If you take a look at Lars Hörmander's ALPDO first volume, Springer Grundlehren 256 on page 300, you will find the definition of the normal set to any closed subset of $\mathbb R^n$ and Theorem 8.5.6' asserts that the analytic wave-front-set of a distribution $u$ contains the normal set to the support.

For the example you give, I believe that the $C^\infty$ wave-front-set is simply the product of the boundary with $\mathbb R^2\backslash {(0,0)}$. However for $s$ real, we may define the $H^s$ wave-front-set and I guess that the Hausdorff dimension of the fractal set you consider should be linked to the $H^s$ wave-front-set, in the sense that it should be trivial at some threshold $s_0$ and that $s_0$ should be linked to the Hausdorff dimension. The paper MR1277392, by Falcone, is addressing a related question.

  • $\begingroup$ hi, What is the name of the paper? $\endgroup$
    – yess
    Commented Apr 7, 2023 at 21:43

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