Let $\Omega \subset \mathbb R^n$ be an open bounded domain, $u \colon \Omega \to \mathbb R$ be a Lipshitz function and suppose that $\nabla u (x) \neq 0$ for $x \in \Omega$. The coarea formula tells us that for any measurable bounded $f \colon \Omega \to \mathbb R$ we have $$ \int_\Omega f(x) \, dx = \int_{\mathbb R} \int_{u^{-1}(t)} f(x) \, \frac{dH^{n-1}(x)}{|\nabla u(x)|}\,dt, $$ where $dH^{n-1}$ is the Hausdorff measure. My question is whether this formula remains valid for $u \in C^\infty(\Omega)$ (with bounded derivatives) but with $f(x) = \delta(u(x))$? Is it true, in particular, that for $g \in C^\infty_c(\Omega)$ the following sequence of equalities holds? $$ \int_{u^{-1}(0)} g(x) \frac{dH^{n-1}(x)}{|\nabla u(x)|}=\int_{\mathbb R} \delta(t) \int_{u^{-1}(t)}g(x) \frac{dH^{n-1}}{|\nabla u(x)|} dt \\ = \frac{1}{2\pi}\int_{\mathbb R}\int_{\mathbb R} e^{its}\int_{u^{-1}(t)}g(x) \frac{dH^{n-1}(x)}{|\nabla u(x)|} ds dt \\ = \frac{1}{2\pi} \int_{\mathbb R} \int_\Omega e^{isu(x)} g(x) \, dx ds \quad ? $$
1 Answers
The first equality is true for $u \in C^1(\Omega)$ for which the gradient does not vanish in $\Omega$. You can approximate $ \delta(u(x)) $ with a rescaled bump function $ \rho_\epsilon(x) = \epsilon^{-(n-1)} \rho(x/\epsilon)$ where: $$ \rho \in C^\infty_c(\mathbb R), \, \int_{\mathbb R^n} \rho(x) \,dx = 1 , \, \rho(x) \geq 0. $$
We have: $$ \int_{\mathbb R^n}\rho_\epsilon(u(x)) \, g(x) \, dx = \int_{\mathbb R} \int_{u^{-1}(t)}\rho_\epsilon(u(x)) \, g(x) \frac{d\mathcal H^{n-1}(x)}{\lvert \nabla u(x)\rvert} \, dx \, dt $$ taking the limit for $\epsilon \to 0$, we obtain $$ \int_{\mathbb R^n}\delta(u(x)) \, g(x) \, dx = \int_{\mathbb R}\int_{u(x) = t}\delta(u(x)) \, g(x) \frac{d\mathcal H^{n-1}(x)}{\lvert \nabla u(x)\rvert} \, dx \, dt $$ The right hand side can be written as $$ \int_{\mathbb R} \delta(t) \left( \int_{u(x)=t} \, g(x) \frac{d\mathcal H^{n-1}(x)}{\lvert \nabla u(x)\rvert} \, dx \right)\, dt = \langle \delta_0 \, ; \, F(t) \rangle = F(0) $$ where $$ F(0) = \int_{u(x)=0} \, g(x) \frac{d\mathcal H^{n-1}(x)}{\lvert \nabla u(x)\rvert} \, dx. $$ For the other equalities I guess you are writing $\delta$ as the inverse Fourier transform (on distribution) of the constant function $1$, which is fine when you are integrating against Schwartz functions. The above calculation makes sense when $F(t)$ is just continuous (to apply the Dirac Delta $\delta$).
Even in one dimension, on $\Omega = (0,1)$ the function $u(x) = x^2$ is smooth with bounded derivatives on $\Omega$, but $1/(u'(x))$ is not bounded on $(0,1)$.
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