composition lemma

[tommy1729]

Let f [ g^(-1)[x] ] + g^(-1) [ f[x] ] = 2 g[x]

replace x with g(x)

hence we get for x e image g(x)

f(x) + g^(-1)[ f[g(x)] ] = 2 g[g(x)]

if f and g are real analytic then for all real x

f(x) + g^(-1)[ f[g(x)] ] = 2 g[g(x)]

Let g(g(x)) = f(x) + r(x)/2

hence

f(x) + g^(-1)[ f[g(x)] ] = 2 f(x) + r(x)

g^(-1)[ f(g(x)) ] = f(x) + r(x)

thus r(x) = 0 for all x iff f(g(x)) = g(f(x))

which is the condition we needed.

QED

tommy1729