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For $c_1,\dots,c_n\in \mathbb{R}^l$ fixed, I am trying to show that

If $0\in int \{\sum_{i=1}^n\alpha_ic_i:\alpha_i\ge 0,\ \sum_{i=1}^n \alpha_i=1\}$, then $0\in int \{\sum_{i=1}^n\alpha_iv_{i}:\alpha_i\ge 0,\ \sum_{i=1}^n \alpha_i=1\}$ for all $v_{1},\dots,v_{n}\in \mathbb{R}^l$ with $ \sum_{i=1}^n||c_i-v_{i}||<\varepsilon$, if $\varepsilon$ is small enough.

Would appreciate some help

2 Answers 2

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Let $C_0 = \{c_1, \dots, c_n\}$, and let $C$ be the convex hull of $C_0$ in $\mathbf{R}^l$. Since $0 \in \operatorname{int} C$, there is some simplex $D$ with vertices $d_0,\dots, d_l$ all in $C$ such that $0$ is in the interior of $D$. (For example, $D$ is a tetrahedron when $l = 3$.)

The point $0$ can be written uniquely as a convex combination $\sum \beta_j d_j$ with $\beta_j > 0$.

Now write each $d_j$ as a convex combination $\sum_i \alpha_{j,i} c_i$, and let $w_j = \sum_i \alpha_{j,i} v_i$. As $v_i \to c_i$, we have $w_j \to d_j$. For $v_i$ close enough to $c_i$, the $w_j$'s will be affinely independent since the $d_j$'s are (i.e., $\det(w_1-w_0,\dots,w_l-w_0)\ne 0$).

Now write $0$ (uniquely) as an affine combination $\sum \gamma_j w_j$. Since $w_j \to d_j$ and the coefficients $\gamma_j$ are continuous functions of the $w_j$'s, we have $\gamma_j \to \beta_j >0$. Once all the $\gamma_j$'s are positive, $0$ will be in the interior of the simplex generated by the $w_j$'s.

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    I still do not get the completly idea of... "For $v_1$ close enough to $c_i$, the $w_i$`s will be affinely independent since the $d_j$`s are" – 2017-02-22
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    In general, I want to understand the Lemma: $u_1,u_2,\ldots, u_n$ are L i and $v_1,v_2,\ldots, v_n$ are "very near" $u$´s implies $v_1,v_2,\ldots, v_n$ are L i – 2017-02-22
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    By L. i., I think you mean "linearly independent". The set of $n$-tuples $(v_1,\dots,v_n)$ such that $v_1,\dots,v_n$ are linearly independent is open, since it is the inverse image of $\mathbf{R} -\{0\}$ under the continuous mapping $(v_1,\dots,v_n) \mapsto \det(v_1,\dots,v_n)$. If you assume that $(u_1,\dots,u_n)$ is in the set, it must be an interior point of the set. – 2017-02-22
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    Likewise, $w_0, \dots, w_l$ are affinely independent if and only if $\det(w_1-w_0,\dots,w_l-w_0) \ne 0$. But this is a continuous function of $(v_1, \dots, v_n)$. – 2017-02-22
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    And in the last part there are many ideas that are not very clear: Why can you express $0$ as an affine combination ?, how did you find $\gamma_j$´s?, why are the coefficients $\gamma_j$´s continuous functions of the $w_j$'s? – 2017-02-23
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    why do you conclude directly that $0$ is in the interior of the simplex generated by the $w_j$'s? – 2017-02-23
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    The points $w_0, \dots, w_l$ being affinely independent, every point in $\mathbf{R}^l$ can be written as $\sum \gamma_j w_j$ with $\sum \gamma_j = 1$ in a unique way. This correspondence constitutes an affine bijection between $\mathbf{R}^l$ and the hyperplane $\gamma_0 + \dots + \gamma_l = 1$ of $\mathbf{R}^{l+1}$. Under this correspondence, the simplex generated by the $w_j$'s corresponds to the "standard" simplex $\sum \gamma_j = 1, \gamma_j \geq 0$. The interior of the simplex generated by the $w_j's$ corresponds to the interior of the standard simplex, which is defined... – 2017-02-23
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    by $\gamma_j > 0$. I'll write the proof of continuity later. – 2017-02-23
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    The equations $\sum \gamma_j = 1, \sum \gamma_j w_j = 0$ constitute a system of $l + 1$ linear equations in the $l+1$ unknowns $\gamma_j$. We have said that there is a unique solution. It follows that the unknown coefficients $\gamma_j$ can be obtained by Cramer's Rule, and are therefore continuous functions of the $w_j$'s. – 2017-02-23
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Let I be a closed and bounded interval, and f (x): I-> I is a continuous function. Show that there exists x element of I such that f (x) = x proves that x is not necessarily unique