Prove that f(x) is constant...?

• April 25th 2009, 10:19 PM
fardeen_gen
Prove that f(x) is constant...?
Suppose that function f : R R satisfies the inequality:
n
| 3^r{f(x + ry) - f(x - ry)}|
r = 1
Prove that f(x) = constant

• May 27th 2009, 02:51 PM
Media_Man
Clarity?
If $abs(\sum_{r=1}^n 3^r[f(x+ry)-f(x-ry)])\leq 1$ for all $n\in\mathbb{N}$, $x,y\in\mathbb{R}$ then f is constant.

Suppose f is not constant. Then there exists an a,b such that $f(a)\neq f(b)$, or $f(a)=f(b)+\epsilon$ for some $\epsilon\neq 0$. Choose a positive integer n such that $3^n|\epsilon|>2$. Define x,y such that $a=x+ny$ and $b=x-ny$

Now, by construction, $|a_n|>2$, for $a_n$ the last term of the summation. Now we have two cases:

Case 1: $abs(\sum_{r=1}^n a_r)>1$ so f does not meet the hypothesis for the choice n,x,y as prescribed.

Case 2: $abs(\sum_{r=1}^n a_r)\leq1$ or, the terms $a_r$ for $r counterbalanced $a_n$ to put the whole sum back under 1.

More rigorously, $\sum_{r=1}^n a_r\in[-1,1]$. WLOG, suppose $|a_n|=a_n>2$. Then $\sum_{r=1}^{n-1} a_r = \sum_{r=1}^n a_r - a_n<-1$. So f does not meet the hypothesis for the choice n-1,x,y as prescribed.

Ergo, by contrapositive, the theorem is true.