Smooth boundary condition implies exterior sphere condition

If $\Omega$ is a bounded domain in $\mathbf{R}^n$ with $C^2$ boundary, show that $\Omega$ satisfies "exterior sphere condition".

Exterior sphere condition means that for each $z\in\partial\Omega$, there is a ball $B_r(\xi)$ satisfying $\overline{B_r(\xi)}\cap\overline{\Omega}=\{z\}$.

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Could you provide context about why you are interested in this? – Grumpy Parsnip Jun 4 '11 at 12:20

Given a point on the boundary of $\Omega$, choose an orthogonal coordinate system $x_1,\ldots,x_n$ for $\mathbb{R}^n$ so that the point lies at $(0,0,\ldots,0)$ and the tangent hyperplane to $\partial\Omega$ at the point is the hyperplane $x_n = 0$.
In this case, by the Implicit Function Theorem, there is an $\epsilon$-neighborhood of the origin on which $\Omega$ is defined by $$x_n \;<\; f(x_1,\ldots,x_{n-1})$$ where $f\colon\mathbb{R}^{n-1}\to\mathbb{R}$ is a $C^2$ function and $f(0,\ldots,0)=0$. This function has a Hessian matrix $Hf$ at the origin, where $$(Hf)_{ij} \;=\; \frac{\partial f}{\partial x_i\partial x_j}(0,\ldots,0).$$ This matrix is symmetric, and hence diagonalizable, with real eigenvalues $\lambda_1,\ldots,\lambda_n$. (These are the principal curvatures of $\partial\Omega$ at the origin.) Choose any radius $0<r<\epsilon/2$ satisfying $$\frac{1}{r} \;\geq\; \max\{\lambda_1,\ldots,\lambda_n\}.$$ Then the closed ball of radius $r$ centered at the point $(0,\ldots,0,r)$ will intersect $\overline{\Omega}$ only at the origin.
Sorry, but why the condition $\frac{1}{r}\ge\max(\lambda_1,\ldots,\lambda_n)$ ensures that the closed ball intersects $\overline{\Omega}$ only at the origin? – NGY Jun 6 '11 at 5:01