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Homework Statement
Let [itex]f: [0,2]\to\mathbb{R}[/itex] be continuous and [itex]f(0) = f(2)[/itex]. Show that there exist [itex]x,y\in [0,2][/itex] with the following property:
[itex](*)\ y-x = 1[/itex] and [itex]f(x) = f(y)\ (*)[/itex]
Homework Equations
Bolzano-Cauchy theorem: If a function [itex]f[/itex] is continuous in some interval [itex][a,b][/itex] and [itex]f(a) <0, f(b) > 0[/itex] (or vice versa) then there exists [itex]c\in (a,b)[/itex] such that [itex]f(c) = 0[/itex]
The Attempt at a Solution
If [itex]f[/itex] was constant, then it's trivial. Fix [itex]x\in [0,1][/itex] and the condition [itex]f(x+1)-f(x) = 0[/itex] is satisfied.
Hence, assume [itex]f[/itex] is not constant.
Let us observe function [itex]g(x) := f(x+1)-f(x)[/itex], [itex]0\leq x\leq 1[/itex]. The objective is to show that [itex]g(x)=0[/itex] is possible with which we will have proven the existence of the required [itex]x,y[/itex] (is this correct to say? )
Let us note that:
[itex]g(0) = f(1) - f(0)[/itex] and [itex]g(1) = f(2) - f(1) = f(0) - f(1)[/itex]. If [itex]g(0) > 0[/itex], then [itex]g(1) <0[/itex] (or vice versa), we can therefore conclude that:
Per Bolzano-Cauchy theorem there exists [itex]c\in (0,1)[/itex] such that [itex]g(c) = f(c+1) - f(c) = 0[/itex] from which we can establish [itex]x = c[/itex] and [itex]y = c+1[/itex] and the condition [itex](*)[/itex] is satisfied
If [itex]g(0) = 0[/itex] then also [itex]g(1) = 0[/itex] and again the condition [itex](*)[/itex] is satisfied. [itex]Q.E.D[/itex]