## Appendix B  
This Appendix proves some auxiliary results useful for proving Theorems 2-3. Henceforth, \(\|x\|_{2}\) shall denote the Euclidean norm of a vector \(x\) in a finite-dimensional Euclidean space. **Lemma B.1. ** Let \(\mathcal{T}\) be an IFS of contraction mappings on \(\Omega_{m}\). Let \(\pi\) be the projection mapping \((p,y)\to p\) from \(\Omega_{m}\) onto \(\mathbb{P}_{m}\). Suppose for each \(T\in\mathcal{T}\), there is a contraction mapping \(T^{*}\) on \(\mathbb{P}_{m}\) such that \(T^{*}(p)=\pi(T(p,y))\) for every \((p,y)\) in \(\Omega_{m}\), and suppose \(\mathbb{P}_{m}\) is the attractor of the IFS \(\{T^{*}:T\in\mathcal{T}\}\). Suppose \(h:\mathbb{P}_{m}\to\mathbb{R}\) is a continuous mapping whose graph \(G_{h}=\{(p,h(p)):p\in\mathbb{P}_{m}\}\) satisfies the property  
\[T(G_{h})\subset G_{h},\ \ T\in\mathcal{T}\]  
Then \(G_{h}\) is the attractor of \(\mathcal{T}\). _Proof._ Let \(Q\) be the attractor of \(\mathcal{T}\). Since each mapping in \(\mathcal{T}\) maps the compact set \(G_{h}\) into itself, \(Q\subset G_{h}\) by uniqueness of the attractor. The proof is completed by showing the reverse inclusion \(G_{h}\subset Q\). Since \(\pi(Q)\) is the attractor of the IFS \(\{T^{*}:T\in\mathcal{T}\}\), we must have \(\pi(Q)=\mathbb{P}_{m}\) by assumption. Let \((p,h(p))\) be an arbitrary element of \(G_{h}\). Since \(\pi(Q)=\mathbb{P}_{m}\), there exists a point in \(Q\) of the form \((p,y)\). But \((p,y)\) and \((p,h(p))\) both belong to \(G_{h}\), so \(y=h(p)\). We conclude \((p,h(p))\) belongs to \(Q\), and therefore \(G_{h}\subset Q\). **Lemma B.2. ** Let \(T^{*}:\mathbb{P}_{m}\to\mathbb{P}_{m}\) be a contraction mapping with contraction coefficient \(\sigma\in(0,1)\), meaning that  
\[\|T^{*}(p_{1})-T^{*}(p_{2})\|_{2}\leq\sigma\|p_{1}-p_{2}\|_{2},\ \ p_{1},p_{2}\in \mathbb{P}_{m}\]Let \(c=(c_{1},c_{2},\cdots,c_{m})\) be a vector in \(\mathbb{R}^{m}\) and define its variance by  
\[V(c)\stackrel{{\Delta}}{{=}}\sum_{i=1}^{m}(c_{i}-\bar{c})^{2} \tag{40}\]  
where \(\bar{c}\) is the average of the entries of \(c\). Let \(T:\Omega_{m}\rightarrow\Omega_{m}\) be the mapping  
\[T(p,y)\stackrel{{\Delta}}{{=}}(T^{*}(p),\sigma(y+p\cdot c)),\ \ p\in \mathbb{P}_{m},\ y\in\mathbb{R}\]  
Then \(T\) is a contraction mapping if \(V(c)<\sigma^{-2}(1-\sigma^{2})^{2}\). _Proof._ By the intermediate value theorem, there is a real number \(\lambda\) in the interval \([\sigma,1)\) such that  
\[V(c)=\lambda^{-2}\sigma^{-2}(\lambda^{2}-\sigma^{2})^{2} \tag{41}\]  
Then \(T\) is a contraction if we show that  
\[\|T(p,u)-T(q,v)\|_{2}^{2}\leq\lambda^{2}\|(p,u)-(q,v)\|_{2}^{2} \tag{42}\]  
for \(p,q\) in \(\mathbb{P}_{m}\) and \(u,v\in\mathbb{R}\). The left hand side of Inequality (42) is less than or equal to  
\[\sigma^{2}\|p-q\|_{2}^{2}+[\sigma(p-q)\cdot c+\sigma(u-v)]^{2}\]  
The right hand side of Inequality (42) is equal to  
\[\lambda^{2}\|p-q\|_{2}^{2}+\lambda^{2}(u-v)^{2}\]  
Therefore, we will be done if we can show that  
\[(\lambda^{2}-\sigma^{2})\|p-q\|_{2}^{2}+\lambda^{2}t^{2}-[\sigma(p-q)\cdot c+ \sigma t]^{2}\geq 0 \tag{43}\]  
for all \(p,q\) in \(\mathbb{P}_{m}\) and all real numbers \(t\). If \(V(c)=0\), then we are done because the left side of Inequality (43) is identically zero (this is because \(\lambda=\sigma\) and because \((p-q)\cdot c=0\) due to the fact that the components of \(c\) are constant). We assume \(V(c)>0\) and therefore \(\lambda>\sigma\). Letting \(Q_{p,q}(t)\) for fixed \(p,q\) be the quadratic polynomial  
\[Q_{p,q}(t)=\lambda^{2}t^{2}-[\sigma(p-q)\cdot c+\sigma t]^{2}\]  
the plot of \(Q_{p,q}(t)\) is a parabola opening upward because the coefficient of \(t^{2}\) is the positive number \(\lambda^{2}-\sigma^{2}\).