Estimate the extension of a function on hypersurface

Question

Suppose there is some $(n-1)$-dimensional hypersurface $\Sigma \subset \mathbb{R}^n$. and a smooth function $f:\Sigma \to \mathbb{R}$ satisfying $$|\nabla_{\Sigma} f| \leq C, \quad |\nabla_{\Sigma}^2 f| \leq C^2. $$ Let $\nu(x)$ be a choice of normal and suppose the map $$ F: \Sigma \times (-\epsilon,\epsilon) \to \mathbb{R}^n , \quad (x,s) \to x + s \nu(x)=:\gamma_x(s) $$ is a diffeomorphism for $|s| \leq \epsilon$. I want to extend $f$ to a function $$\bar{f}: \{ y \in \mathbb{R}^n: \inf_{y \in \Sigma} |x-y|\leq \epsilon\} \to \mathbb{R}$$ on a tube around $\Sigma$. My idea was to set $$ \bar{f}(y):= f(y- d(y,\Sigma) \nu(x)). $$ Then $\bar{f}$ stays constant along the geodesic $\gamma_x(s)$, i.e. $$ D_{\nu}\bar{f}\equiv 0. $$ Is there a way to bound the gradient of $\bar{f}$ in some way like $$ |D \bar{f}| \leq C(n) ( |\nabla_{\Sigma} f| + |A|), $$ where $A$ is the second fundamental form of $\Sigma$?

Answer 1

There may be some mistakes in the following, but I think it is okay. We do not have such an additive bound on the norm of $\tilde{f}$ but a more complicated bound.

First, consider the map $E : \Sigma \times (-\varepsilon,\varepsilon) \to \mathbb{R}^n$ defined by $E(p,t) = p + t \nu(p)$, and suppose it is a diffeomorphism onto its image. We denote by $S$ the Weingarten map (or shape operator) of $\Sigma$, that is $S(p) = \mathrm{d}\nu(p)$, the differential of the unit normal. We suppose that $\sup_{p\in \Sigma}\|S(p)\| < + \infty$ (which is true if, for example, $\Sigma$ is compact) where $\|S(p)\|$ denotes the operator norm of $S(p)$, subordinate to the euclidean norm.

Then if $(v,s)$ is a tangent vector to $\Sigma \times (-\varepsilon,\varepsilon)$, at a point $(p,t)$, we have, from the definition of $E$: $$ \mathrm{d}E(p,t)\cdot (v,s) = v + tS(p)v + s \nu(p). $$ Let $f : \Sigma \to \mathbb{R}$ be a smooth function. Extend $f$ on $U = E(\Sigma \times (-\varepsilon,\varepsilon))$ by: \begin{align} \tilde{f}(q) = f(\pi_1\circ E^{-1} (q)), \end{align} where $\pi_1 \colon \Sigma \times (-\varepsilon,\varepsilon)\to \Sigma$ is the projection onto the first factor. We want to bound the differential of $\tilde{f}$. Let us first bound the differential of $\pi_1 \circ E^{-1}$. Write $E^{-1}(q) = \left(\pi_1\circ E^{-1}(q), d(q,\Sigma) \right)=(p,t)$, and note that: $$ \mathrm{d}E^{-1}(q) = \mathrm{d}E\left(E^{-1}(q) \right)^{-1} = \mathrm{d}E\left(p,t \right)^{-1}. $$ Now, note that $\pi_1$ "kills" the normal direction, an hence: $$ \mathrm{d}\left(\pi_1\circ E^{-1}\right)(q) = \left(\mathrm{Id} + tS(p) \right)^{-1}, $$ where we still write $E^{-1}(q) = (p,t)$. Suppose $\varepsilon$ is small enough so that $A = \varepsilon \times \sup_{p\in \Sigma}\|S(p)\| <1$. Then: $$ \left\|\mathrm{d}\left(\pi_1\circ E^{-1} \right)(q) \right\| = \left\|\left(\mathrm{Id} + t S(p)\right)^{-1} \right\| \leqslant \dfrac{1}{1 - |t|\|S(p)\|} \leqslant \frac{1}{1-A}. $$ It follows that: $$ \|\mathrm{d}\tilde{f}(q)\| = \left\|\mathrm{d}f((\pi_1\circ E^{-1})(q) \circ \mathrm{d}\left(\pi_1\circ E^{-1} \right)(q)\right\| \leqslant \|\mathrm{d}f((\pi_1\circ E^{-1})(q))\| \times \frac{1}{1-A} $$ and one can bound $\|\mathrm{d}\tilde{f}\|$ in terms of bounds for $\|\mathrm{d}f\|$ (or equivalently with a bound on $\|\nabla f\|$) and $A$, which depends on the second fundamental form of $\Sigma$. A less brutal bound is: $$ \left\|\mathrm{d}\tilde{f}(p+t\nu(p))\right\| \leqslant \left\|\mathrm{d}f(p)\right\| \times \frac{1}{1 - |t|\|S(p)\|}. $$