| Solution by Chris Wildhagen, Rotterdam, The Netherlands.
Applying f to both sides of the functional equation
f(f(f(x))) = f(x) >= 0 (1)
gives g(g(x)) = g(x), where g(x) = f(f(x)) for all x in R. Of course g
is also a differentiable function on R and g(x) >= 0 for all x in R.
Then the range T = g(r) of g is an interval contained in [0, infinity).
Let a be the infimum of T. Since g(t) = t for all t in T and g is
continuous, it follows that g(a) = a. Assuming that T has more than
one element, choose d > = such that (a, a+d) is a subset of T. Then x
in (a-d, a) implies g(x) >= g(a) (=a), hence
[g(x)-g(a)] / [x-a] <= 0.
Therefore
g'[L](a) = lim as x->a- of[g(x)-g(a)]/[x-a] <= 0. (2)
[The L is brackets above is a subscript in the original. -- edp]
For x in (a, a+d) we have
[g(x)-g(a)] / [x-a] = 1.
Therefore
g'[R](a) = lim as x->a+ of[g(x)-g(a)]/[x-a] = 1. (3)
[The R is brackets above is a subscript in the original. -- edp]
(2) and (3) are contradictory since g is differentiable at a. We are
led to the conclusion that T is a single point, i.e., g is a constant
function, say g(x) = c for all x in R. This gives, using (1), that
f(c) = f(x) for all x in R, showing that f is a constant function.
Thus there is no nonconstant differentiable function satisfying (1).
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