设有 n+1 (n∈N,n≥1)n+1 (n∈N,n≥1) 个点 { αi,i∈N,1≤i≤n+1}{ αi,i∈N,1≤i≤n+1} , βi=αi−αn+1,1≤i≤n.βi=αi−αn+1,1≤i≤n.
引理一
{ ∑n+1i=1xiαi:∑n+1i=1xi=1}={ ∑ni=1xiβi+αn+1,xi∈R}{ ∑i=1n+1xiαi:∑i=1n+1xi=1}={ ∑i=1nxiβi+αn+1,xi∈R}
证明:
由 ∑n+1i=1xi=1⇔xn+1=1−∑ni=1xi∑i=1n+1xi=1⇔xn+1=1−∑i=1nxi 得:
∑n+1i=1xiαi∑i=1n+1xiαi
=∑ni=1xiαi+xn+1αn+1=∑i=1nxiαi+xn+1αn+1
=∑ni=1xiαi+(1−∑ni=1xi)αn+1=∑i=1nxiαi+(1−∑i=1nxi)αn+1
=∑ni=1xi(αi−αn+1)+αn+1=∑i=1nxi(αi−αn+1)+αn+1
=∑ni=1xiβi+αn+1=∑i=1nxiβi+αn+1
反之亦然。
引理二
{ ∑n+1i=1xi<