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Consider two compact convex sets $C_1, C_2 \subset \mathbb{R}^n$ such that $C_2 \subset C_1$. Let us denote by $\partial C_1$ and $\partial C_2$ their boundaries, that satisfy and $\partial C_1 \cap \partial C_2 = \varnothing$.

Consider two continuous, bounded, functions $f_1: C_1 \rightarrow \mathbb{R}^n$ and $f_2: C_1 \rightarrow \mathbb{R}^n$.

Consider a continuous, bounded, function $f: C_1 \rightarrow \mathbb{R}^n$ such that: $$f(y) = f_1(y) \ \ \forall y \in \partial C_1$$

$$f(y) = f_2(y) \ \ \forall y \in \partial C_2$$

1) Prove that there exists a continuous function $g: C_1 \rightarrow \mathbb{R}_{\geq 0}$ such that:

$$ g(y) = 0 \ \ \forall y \in \partial C_1 $$

$$ g(y) = 1 \ \ \forall y \in \partial C_2 $$

$$ f(x) = ( 1-g(x) ) f_1(x) + g(x) f_2(x) \ \ \forall x \in \text{closure}(C_1 \setminus C_2) $$

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You should additionally assume that $\partial C_1$ and $\partial C_2$ are disjoint. (Otherwise, there is no such $g$.) –  Yury Nov 14 '12 at 14:15
    
Can you please explain how $\partial C_1$ and $\partial C_2$ can be "not disjoint" according to $C_2 \subset C_1$? –  Adam Nov 14 '12 at 14:16
1  
For example, $C_1 = C_2 = [0,1]$. Then $\partial C_1 = \partial C_2 = \{0,1\}$. If you mean $C_2\subsetneq C_1$ then we can have $C_2=[0,1/2]$ and $C_1 = [0,1]$. Then $\partial C_2 = \{0,1/2\}$ and $\partial C_2 = \{0,1\}$ are not disjoint. –  Yury Nov 14 '12 at 14:25
    
Ok, I updated the question. So have you got a proof in the disjoint case? –  Adam Nov 14 '12 at 14:32
    
I also inserted convexity for simplicity. –  Adam Nov 14 '12 at 14:34

1 Answer 1

Here is a hint. Let $\rho = d(\partial C_1, \partial C_2) \equiv \min \{\|x_1-x_2\|_2: x_1 \in \partial C_1, x_2\in \partial C_2\}$. Define $h(y) = d(y, C_2) = \min \{\|x-y\|_2: x\in \partial C_2\}$. Now let $g(y) = h(y)/\rho$ if $h(y) \leq \rho$, and $g(y) = 1$, otherwise.

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Why any function $f$ with the given properties can be necessarily written as $f(y) = h(y)/\rho$ if $h(y)\leq \rho$, $1$ otherwise? Isn't function $f$ vector valued? Yours sounds scalar valued. Did you mean $g$ in place of your $f$? –  Adam Nov 14 '12 at 15:34
    
Yes, I meant $g$. I fixed my answer. –  Yury Nov 14 '12 at 16:02
    
Do you mean $h(y) = d(y,\partial C_1)$? –  Adam Nov 15 '12 at 11:33
    
Moreover, why the third property we are seeking ($f = (1-g)f_1 + g f_2$) should be necessarily satisfied with your particular $g$? –  Adam Nov 15 '12 at 11:37

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