210 CHAPTER 10. BROUWER FIXED POINT THEOREM Rn∗
If cn = 0, then by assumption, each c j = 0. The proof goes by assuming cn ̸= 0 and derivinga contradiction. Assume then that cn ̸= 0. Then you can divide by it and obtain modifiedconstants, still denoted as c j such that
b =1
n+1
n
∑i=0
xi = y0 +n−1
∑j=1
c j (y j−y0)
Thus
1n+1
n
∑i=0
∑s ̸=k
t0s (xi−xs) =
n−1
∑j=1
c j (y j−y0) =n−1
∑j=1
c j
(∑s ̸=k
t js xs−∑
s ̸=kt0s xs
)
=n−1
∑j=1
c j
(∑s ̸=k
t js (xs−x0)−∑
s ̸=kt0s (xs−x0)
)Modify the term on the left and simplify on the right to get
1n+1
n
∑i=0
∑s ̸=k
t0s ((xi−x0)+(x0−xs)) =
n−1
∑j=1
c j
(∑s ̸=k
(t js − t0
s)(xs−x0)
)
Thus,
1n+1
n
∑i=0
(∑s ̸=k
t0s
)(xi−x0) =
1n+1
n
∑i=0
∑s ̸=k
t0s (xs−x0)
+n−1
∑j=1
c j
(∑s ̸=k
(t js − t0
s)(xs−x0)
)
Then, taking out the i = k term on the left yields
1n+1
(∑s ̸=k
t0s
)(xk−x0) =−
1n+1 ∑
i̸=k
(∑s ̸=k
t0s
)(xi−x0)
1n+1
n
∑i=0
∑s̸=k
t0s (xs−x0)+
n−1
∑j=1
c j
(∑s ̸=k
(t js − t0
s)(xs−x0)
)That on the right is a linear combination of vectors (xr−x0) for r ̸= k so by independence,∑r ̸=k t0
r = 0. However, each t0r ≥ 0 and these sum to 1 so this is impossible. Hence cn = 0
after all and so each c j = 0. Thus [y0, · · · ,yn−1,b] is an n simplex.Now in general, if you have an n simplex [x0, · · · ,xn] , its diameter is the maximum of
|xk−xl | for all k ̸= l. Consider∣∣b−x j
∣∣ . It equals∣∣∣∣∣ n
∑i=0
1n+1
(xi−x j)
∣∣∣∣∣=∣∣∣∣∣∑i̸= j
1n+1
(xi−x j)
∣∣∣∣∣≤ nn+1
diam(S)