Affine-linear mapping/Barycentric combination/Fact/Proof

Let ${\displaystyle {}V}$ and ${\displaystyle {}W}$ denote the vector spaces for ${\displaystyle {}E}$ and for ${\displaystyle {}F}$, respectively. Suppose first that ${\displaystyle {}\psi }$ is affine-linear with linear part

${\displaystyle \psi _{0}\colon V\longrightarrow W.}$

Let a barycentric combination ${\displaystyle {}\sum _{i\in I}a_{i}P_{i}}$ with ${\displaystyle {}P_{i}\in E}$ and ${\displaystyle {}\sum _{i\in I}a_{i}=1}$ be given. Then we have (with an arbitrary point ${\displaystyle {}Q\in E}$)

${\displaystyle {}{\begin{aligned}\psi {\left(\sum _{i\in I}a_{i}P_{i}\right)}&=\psi {\left(Q+\sum _{i\in I}a_{i}{\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\psi _{0}{\left(\sum _{i\in I}a_{i}{\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\sum _{i\in I}a_{i}\psi _{0}{\left({\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\sum _{i\in I}a_{i}{\overrightarrow {\psi (Q)\psi (P_{i})}}\\&=\sum _{i\in I}a_{i}\psi (P_{i}).\end{aligned}}}$

Now, suppose that the mapping ${\displaystyle {}\psi }$ is compatible with barycentric combinations. We set

${\displaystyle {}\varphi (v):={\overrightarrow {\psi (P)\psi (P+v)}}\,}$

for ${\displaystyle {}v\in V}$, where ${\displaystyle {}P\in E}$ is any point. We first show that this is independent of the chosen point ${\displaystyle {}P}$. For another point ${\displaystyle {}Q\in E}$, the sum

${\displaystyle (P+v)-(Q+v)+Q}$

is a barycentric combination of the point ${\displaystyle {}P}$, see exercise. Therefore, we have in ${\displaystyle {}F}$ the equality

${\displaystyle {}\psi (P+v)-\psi (Q+v)+\psi (Q)=\psi (P)\,.}$

Hence, we have in ${\displaystyle {}V}$ the equality

${\displaystyle {}{\overrightarrow {\psi (P)\psi (P+v)}}-{\overrightarrow {\psi (P)\psi (Q+v)}}+{\overrightarrow {\psi (P)\psi (Q)}}=0\,,}$

and, therefore,

${\displaystyle {}{\begin{aligned}{\overrightarrow {\psi (P)\psi (P+v)}}&={\overrightarrow {\psi (P)\psi (Q+v)}}-{\overrightarrow {\psi (P)\psi (Q)}}\\&={\overrightarrow {\psi (P)\psi (Q+v)}}+{\overrightarrow {\psi (Q)\psi (P)}}\\&={\overrightarrow {\psi (Q)\psi (Q+v)}}.\end{aligned}}}$

We have to show that ${\displaystyle {}\varphi }$ is linear. For ${\displaystyle {}u={\overrightarrow {PQ}}}$ and ${\displaystyle {}v={\overrightarrow {PR}}}$, we have

${\displaystyle {}{\begin{aligned}\psi (P)+\varphi (au+bv)&=\psi (P+au+bv)\\&=\psi {\left(P+a{\overrightarrow {PQ}}+b{\overrightarrow {PR}}\right)}\\&=\psi ((1-a-b)P+aQ+bR)\\&=(1-a-b)\psi (P)+a\psi (Q)+b\psi (R)\\&=\psi (P)+a{\overrightarrow {\psi (P)\psi (Q)}}+b{\overrightarrow {\psi (P)\psi (R)}}\\&=\psi (P)+a\varphi {\left({\overrightarrow {PQ}}\right)}+b\varphi {\left({\overrightarrow {PR}}\right)}\\&=\psi (P)+a\varphi (u)+b\varphi (v).\end{aligned}}}$

Thus, we have

${\displaystyle {}\varphi (au+bv)=a\varphi (u)+b\varphi (v)\,.}$