This is a problem number 8 from ISI BMath 2007 based on the Continuity and composition of a function. Try this out.
Problem: Continuity and composition of a function
Let 
be a continuous function such that P(X)=X has no real solution. Prove that P(P(X))=X has no real solution.
Discussion:
Hunch: There is no solution of P(x) = x implies the graph of P(x) never ‘crosses’ the y=x line,
Suppose P(P(x)) = x has a solution at x = a then P(P(a)) = a.
Suppose P(a) = b (b is not equal to a as P(x) = x has no solution) then P(P(a)) = a implies P(b) = a. Hence we have the following:
P(a) = b and P(b) = a.
There fore the points (a, b) and (b, a) are both on the graph of P(x). But these two points are reflections about y=x line implying P(x) is on both side of that line hence continuity implies P(x) must intersect y=x line leading to a solution of P(x) = x hence contradiction. Therefore P(P(x)) = x has not solution.
Final Solution
Consider the auxiliary function g(x) = P(x) – x. If P(P(x)) = x has a solution then we have already established that there exists a, b such that P(a) = b and P(b) = a. We plug in a, b in g(x).
g(a) = P(a) – a = b – a
g(b) = P(b) – b = a – b
Now one of a-b and b-a is positive. The other is negative. Since P(x) is continuous so is g(x). So by intermediate value property theorem, g(x) will become 0 for some value c between a and b.
g(c) = 0 implies P(c) – c = 0 or P(c) = c a contradiction.
Hence P(P(x)) = x has no solution.
Proved
Some Useful Links:
How to use invariance in Combinatorics – ISI Entrance Problem – Video
Trackbacks/Pingbacks