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Show that the barycentric coordinates $t_o(x), t_1(x),...,t_n(x)$ continually depend on $x$. (Assuming that $a_0,...,a_n$ are geometrically independent)

I'm a little confused with this problem and I'm not sure what I have to prove, I think it's the following:

For all $\epsilon>0$ and $i=0,1,...,n$ there is a $\delta>0$ such that if $|x-y|<\delta$ then $|t_i(x)-t_i(y)|<\epsilon$.

But I do not know how to use the vectors to be geometrically independent to do this, could someone help me please? Thank you very much.

user402543
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    If your points are affinely independant in say $\mathbb{R}^{n}$ then you can find a homoemorphism between the simplex (the convex hull) spanned by the $a_i$'s and the convex hull of ${0, e_1,...,e_n}$ where $(e_i)$ is the canonical basis. The barycentric coordinates are then the projection on the axis, which are continuous – Ahr Jun 05 '18 at 07:53

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Check my answer on same question https://math.stackexchange.com/a/4487873/557871

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Just as comment suggests $t_i(x)$ is same function with composition of $\vec{-a_0}$translation, linear isomorphism by assigning basis of vector space, and projection. Continuity of other functions are trivial, but continuity of isomorphism is may not be trivial. This matter might not interest many but it’s only trivial to those who knows exactly why or those who are used to topological vector space argument. Actually this is more of analysis problem. This can be proved by statement “All linear mapping between finite dimensional Euclidean space is continuous”. Check analysis textbook with multivariable part. Cauchy Schwarz inequality would be key.

For those who are interested in more general continuity matter between any finite dimensional vector space, check this article "Every linear mapping on a finite dimensional space is continuous"

tolmekia
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