# is signed distance function real analytic for real analytic domains

If $$\Omega$$ is a real analytic domain in $$\mathbb R^n$$, is the signed distance function, $$f$$, defined by $$$$f(x)=\begin{cases}d(x,\partial \Omega )&{\mbox{ if }}x\in \Omega \\-d(x,\partial \Omega )&{\mbox{ if }}x\in \Omega ^{c}\end{cases}$$$$ is real analytic? where $$$$d(x,\partial \Omega ):=\inf _{y\in \partial \Omega }d(x,y)$$$$ where inf denotes the infimum.

• The following paper by Lieberman may help answer this question (if you increase boundary regularity in the regularised distance function construction and prove a few more results): msp.org/pjm/1985/117-2/p08.xhtml – JCM Sep 21 '18 at 4:42
• Do you mean, real analytic on a neighborhood of $\partial \Omega$? It surely can't be real analytic everywhere, unless I'm badly misunderstanding definitions. – Nate Eldredge Sep 21 '18 at 5:16
• yes, I mean, is real analytic in a neighborhood of $\partial \Omega$? – mathpde Sep 21 '18 at 12:16

The answer is yes, that is $$f(x)$$ is real analytic in a neighborhood of any point on $$\partial\Omega$$.
First recall that if $$f$$ and $$g$$ are real analytic functions (of several variables), then $$f+g$$, $$f\cdot g$$, $$f/g$$ (when $$g\neq 0$$), $$f\circ g$$, and the inverse map $$f^{-1}$$, if $$f$$ is a diffeomorphism, are all real analytic. You can find proofs in the book [1].
If $$\partial\Omega$$ is locally the image of a real analytic embedding $$\Phi:\mathbb{R}^{n-1}\supset U\to\mathbb{R}^n$$, then $$N(x)$$, the unit normal vector orthogonal to the image of $$D\Phi(x)$$ in the interior direction of $$\Omega$$ is also real analytic. Indeed, $$D\Phi$$ is real analytic and we find a normal vector by solving linear equations involving $$D\Phi(x)$$ so there is a real analytic normal vector $$M(x)$$. Possibly $$M$$ is not unit, but $$N(x)=M(x)/|M(x)|$$ is real analytic, because it is obtained from $$M$$ by applying to $$N$$ operations (listed above) that preserve analyticity.
The mapping $$\Psi:U\times(-\varepsilon,\varepsilon)\to\mathbb{R}^n$$, $$\Psi(x,t)=\Phi(x)+tN(x)$$ is a real analytic diffeomorphism (if $$U$$ and $$\varepsilon$$ are small enough) so the inverse mapping $$\Psi^{-1}:W\to U\times(-\varepsilon,\varepsilon)$$, defined in a neighborhood of a point on $$\partial\Omega$$ is also real analytic. If $$\pi:U\times(-\varepsilon,\varepsilon)\to(-\varepsilon,\varepsilon)$$ is the projection on the $$t$$ component, then $$\pi\circ\Psi^{-1}$$ is real analytic and it remains to observe that the signed distance satisfies $$f(x)=\pi\circ\Psi^{-1} \quad \text{in} \quad W$$ if $$W$$ is small. That follows immediately from the fact that the distance to the boundary is measured (near the boundary) along the normal line.