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The following estimate arises in the proof of Tomas-Stein restriction theorem. $$ \sigma_{\mathbb{S}^{d-1}} (B(x,r)\cap \mathbb{S}^{d-1}) \leq C r^{d-1} $$ The estimate is very intuitive, and I have a vague idea to prove it (essentially estimating the intersected area with the area of the circle of radius $r$ in $\mathbb{R}^{d-1}$ up to global constants, thanks to the compactness of $\mathbb{S}^{d-1}$).

However, when trying to fill in the details, I got stuck, trying to produce a nice and clean proof. In particular, I would love to have a simple and clean proof of this statement possibly extending it to the case where $S$ is a compact $d-1$ dimensional manifold, so any contribution or suggestion is deeply appreciated.

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  • $\begingroup$ What exactly did you try? $\endgroup$ Commented Nov 30, 2025 at 13:27
  • $\begingroup$ Well, the line of reasoning was the following. We find an atlas such that for each chart the derivatives of the coordinate maps are bounded uniformly (this guarantees that up to constants the area of the d-1 dimensional disc of radius r is equivalent to the area of its image in $\mathbb{S}^{d-1}$. The only thing it remains to check is that indeed $\forall r \quad \exists C$ such that $B(x,r) \cap \mathbb{S}^{d-1} \subseteq \phi_i^{-1}(B(0,Cr))$ where $\phi_i$ is a chart of the atlas. $\endgroup$ Commented Nov 30, 2025 at 13:46
  • $\begingroup$ Is your ball $B(x, r)$ a ball in $\mathbb R^d$? I guess I will be much easier to not use the atlas (of $\mathbb S^{d-1}$) and directly calculate using Euclidean geometry $\endgroup$ Commented Nov 30, 2025 at 14:01
  • $\begingroup$ Yes, but I'd like a proof that can be possibly applied to any d-1 dimensional compact manifold $S$. $\endgroup$ Commented Nov 30, 2025 at 14:37
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    $\begingroup$ I’m sure you can extract a proof from parts of my answer to Bound for the surface of a ball lying inside of another ball. There part of the difficulty is in trying to get estimates uniform in $t$ and $R$, but you have $t=1$. $\endgroup$ Commented Nov 30, 2025 at 16:21

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For large $r$ the estimate is trivial so we only need to prove the result for $r$ sufficiently small. We may also assume $x\in S$, as if not, then there is by triangle inequality $C'>0$ and $x'\in S$ such that $B(x',C'r) \supset B(x,r)$, and we can absorb $C'$ into $C$.

So let $x\in S$. Then on a sufficiently small ball $B_{\rho_x}(x)$ around $x$ in $\mathbb R^d$, we can describe (after a rotation if necessary) $B_{\rho_x}(x)\cap S = \{(z,\phi(z)): z\in U_x\}$ as the graph of a nice function $\phi: U_x\to \mathbb R$, $U_x$ open, with some $z_0\in U_x$ such that $x=(z_0,\phi(z_0))$, and $\nabla\phi(z_0)=0$. By taking $\rho_x\ll1$ we may assume $|\nabla\phi(z)|<1$ in $U_x$.

In graph coordinates the surface measure can be written $\sqrt{1+|\nabla\phi(z)|^2} dz$. For any point $y=(z,\phi(z))\in B_{r}(x)\cap S$, $r<\rho_x$, we have $|z-z_0| < r$. Thus

$$\int_{B_{r}(x)\cap S} d\sigma \le \int_{|z-z_0|<r} \sqrt{1+|\nabla\phi(z)|^2} dz\le C r^{d-1}.$$

By compactness there exists a uniform $\rho$ such that the result holds for all $x\in S$ and all $r<\rho$. And we are done.

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