Suppose that function f : R → R satisfies the inequality:
n
| ∑ 3^r{f(x + ry) - f(x - ry)}| ≤ 1 for every positive integer r, x, y
r = 1
Prove that f(x) = constant
If $\displaystyle abs(\sum_{r=1}^n 3^r[f(x+ry)-f(x-ry)])\leq 1$ for all $\displaystyle n\in\mathbb{N}$, $\displaystyle x,y\in\mathbb{R}$ then f is constant.
Suppose f is not constant. Then there exists an a,b such that $\displaystyle f(a)\neq f(b)$, or $\displaystyle f(a)=f(b)+\epsilon$ for some $\displaystyle \epsilon\neq 0$. Choose a positive integer n such that $\displaystyle 3^n|\epsilon|>2$. Define x,y such that $\displaystyle a=x+ny$ and $\displaystyle b=x-ny$
Now, by construction, $\displaystyle |a_n|>2$, for $\displaystyle a_n$ the last term of the summation. Now we have two cases:
Case 1: $\displaystyle abs(\sum_{r=1}^n a_r)>1$ so f does not meet the hypothesis for the choice n,x,y as prescribed.
Case 2: $\displaystyle abs(\sum_{r=1}^n a_r)\leq1$ or, the terms $\displaystyle a_r$ for $\displaystyle r<n$ counterbalanced $\displaystyle a_n$ to put the whole sum back under 1.
More rigorously, $\displaystyle \sum_{r=1}^n a_r\in[-1,1]$. WLOG, suppose $\displaystyle |a_n|=a_n>2$. Then $\displaystyle \sum_{r=1}^{n-1} a_r = \sum_{r=1}^n a_r - a_n<-1$. So f does not meet the hypothesis for the choice n-1,x,y as prescribed.
Ergo, by contrapositive, the theorem is true.