If for any $\epsilon >0$,there exists $a_0$(depends on $\epsilon$) such that $f(a_0)>1-\epsilon$,where $f$ is a real function.We also know that $f(a_0)\leq 1$,can we deduce that $f(a_0)^4=f(a_0)$?
My thought:$1\geq f(a_0)\geq f(a_0)^4 >(1-\epsilon)^4$,then$ |f(a_0)-f(a_0)^4|<1-(1-\epsilon)^4$,the right side of the inequality tends to zero.
Is it correct?
Let a = a$_0$, r = $\epsilon$. As a is dependent
upon r, nothing can be concluded about f(a).
For example, let r = 1. All that is possible
is that exists a with 1 >= f(a) > 0.
Pick a different r, you get a different a with different results,
If r approaches 0, then at each step there is a different a.
When claiming f(a) = f(a)$^4$, what a do you mean?
If a is a constant, it is easy to prove f(a) = 1.