05/27/2011, 07:13 PM
yes that is true.
for those confused :
0 < x f(f({x}/e)) = {x}/e.
reduces to
0 < x < e f(f(x)) = x
in fact i noticed i made a mistake. ( when i had no computer in the neighbourhood )
if f(x) is real-analytic we get a contradiction since f(f(x)) is then also real-analytic and the equation f(f(x)) = x leads to f(f(x)) - x = 0 where f(f(x)) - x is also real-analytic.
but f(f(x)) - x = 0 for 0 < x < e
so on the interval [0,e] we simply have a constant 0 function but another function elsewhere ; this clearly is not real-analytic.
so the question reduces to finding :
non-linear Coo f(x) that satisfies for 0 < x < e => f(f(x)) = x
that should have been the OP.
for those confused :
0 < x f(f({x}/e)) = {x}/e.
reduces to
0 < x < e f(f(x)) = x
in fact i noticed i made a mistake. ( when i had no computer in the neighbourhood )
if f(x) is real-analytic we get a contradiction since f(f(x)) is then also real-analytic and the equation f(f(x)) = x leads to f(f(x)) - x = 0 where f(f(x)) - x is also real-analytic.
but f(f(x)) - x = 0 for 0 < x < e
so on the interval [0,e] we simply have a constant 0 function but another function elsewhere ; this clearly is not real-analytic.
so the question reduces to finding :
non-linear Coo f(x) that satisfies for 0 < x < e => f(f(x)) = x
that should have been the OP.
