2
$\begingroup$

The following statement is excerpted from the proof of a theorem about integral operator.

Let $D\subset{\mathbb R}^m$ be a bounded domain. If $\partial D$ is of class $C^1$, then the normal vector $\nu$ is continuous on $\partial D$. Therefore, we can choose $R\in(0,1]$ such that $ \nu(x)\cdot\nu(y)\geq\frac{1}{2} $ for all $x,y\in\partial D$ with $|x-y|\leq R$.

I can't understand the "therefore" part. Maybe the key point is in the connection between the continuity of $\nu$ and the "$\geq$" inequality. But I don't see any relationship between them. Can any one tell me what's the reasoning there?

2 Answers 2

2

Since $2\nu(x)\cdot\nu(y) =-\lVert \nu(x)-\nu(y)\rVert^2+\lVert \nu(x) \rVert^2+\lVert\nu(y)\rVert^2 =-\lVert \nu(x)-\nu(y)\rVert^2+2$, we have to show that we can find $R\in \left(0,1\right]$ such that $-\lVert \nu(x)-\nu(y)\rVert^2+2\geq 1$ if $\lVert x-y\rVert\leq R$, i.e. $\lVert \nu(x)-\nu(y)\rVert\leq 1$ if $\lVert x-y\rVert\leq R$. Now $\partial D$ is compact as a closed subset of the compact $\overline D$, and $\nu$ is continuous, hence uniformly continuous. Now, just apply the definition of the uniform continuity with $\varepsilon=1$ to get the result.

2

By definition $\nu(x)\cdot \nu(x)=1$, so if $y$ is close to $x$ and $\nu$ is continuous, $\nu(x)\cdot\nu(y)$ will be close to $1$. How close does it need to be? Because $D$ is assumed bounded, $\partial D$ is compact, so the necessary bound $R(x)$ for each $x$ will have a global maximum somewhere.