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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 $g$.)2012-11-14
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    Can you please explain how $\partial C_1$ and $\partial C_2$ can be "not disjoint" according to $C_2 \subset C_1$?2012-11-14
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    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**.2012-11-14
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    Ok, I updated the question. So have you got a proof in the disjoint case?2012-11-14
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    I also inserted convexity for simplicity.2012-11-14
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    You don't want sets $C_1$ and $C_2$ to be disjoint (if they are disjoint then $C_2=\varnothing$); rather sets $\partial C_1$ and $\partial C_2$ should be disjoint. What did you try?2012-11-14
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    You are right... Clearly it was just a typo.2012-11-14
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    If $x$ is in the closure of $C_1\setminus C_2$, it may well be outside $C_2$, hence $f_2(x)$ may not even be defined. Then it what sense shuld we have $f(x)=\ldots +g(x)f_2(x)$?2012-11-14
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    Fixed. In view of the problem statement, there is no substantial reason for not having $f_2$ defined on the whole $C_1$.2012-11-14

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