Projection of a ball in the ambient space to a manifold Let $B_h (x)$ be the ball of radius $0<h \ll 1$ centered at $x\in \mathbb{R}^d$.
Let $I=[0,1]^{d-1}$ be the unit cube in $\mathbb{R}^{d-1}$, and let $f:I \to \mathbb{R}$ be a $C^2$ function. Then $$M:\,=\left\{(x,f(x)) ~ ~ | ~ ~ x\in I \right\} \, ,$$
is a $d-1$ dimensional manifold embedded in $\mathbb{R}^d$.
My questions:


*

*Is $B_h (x) \cap M$ contained in a $d-1$ dimensional ball on $M$ with radius $O(h)$ as $h\to 0$, in the manifold's induced Riemannian metric?

*Let $\sigma $ be the surface-measure on $M$ induced by the Lebesgue measure on $\mathbb{R}^d$. Is $\sigma \left( B_h (x) \cap M\right) = O(h)$ as $h\to 0$? Is it $O(h^{d-1})$?

*If one of the above answers are "no", what do we need to demand on $f$ to amend it? A finite curvature of $M$?


Intuitively, I think that both answers should be yes because (somehow) $M$ should have a bounded curvature. But I wasn't able to prove it. A reference would be also as useful.
(Cross-posted from math.se, after 10 days of virtually no activity)
 A: If the induced Riemannian metric refers to the (Euclidean) length $d(x, y)$ of the shortest path contained in $M$ with given endpoints $x, y$, then this is a completely elementary question. Or, I have misunderstood the problem completely: in this case let me know and I will delete this answer.

Clearly, $d(x, y) \geqslant |x - y|$. On the other hand, set $x = (\tilde{x}, f(\tilde{x}))$ and $y = (\tilde{y}, f(\tilde{y}))$, and define $L = |\tilde{y} - \tilde{x}|$ and $\tilde{z} = (\tilde{y} - \tilde{x}) / L$. We have
$$ d(x, y) \leqslant \int_0^L \sqrt{1 + (D_{\tilde{z}} f(\tilde{x} + t \tilde{z}))^2} dt \leqslant \sqrt{1 + \|\nabla f\|_\infty^2} \times L,$$
where $D_{\tilde{z}}$ is the directional derivative. Therefore,
$$ |x - y| \leqslant d(x, y) \leqslant \sqrt{1 + \|\nabla f\|_\infty^2} \times |x - y| . $$
This gives an affirmative answer to question 1.
Similarly, the volume element of $M$ is given by $$\sqrt{1 + |\nabla f(\tilde{x})|^2} d\tilde{x},$$ and so it is uniformly comparable with $d\tilde{x}$. It follows that the answer to question 2 is also 'yes'.
