Re: F(F(x))+2F(x)=3x, existence of fixed points

Partial answer:

Let B be a Banach space, F:B-->B be a continuous mapping satisfying the functional equation F(F(x))+2F(x)=3x for all x in B. Assume that the range of F is dense in B.
Then F has at least one fixed point.

Proof.

Since F is continuous, for each a in B and eps>0 there is
delta=delta(a,eps)>0 such that ||x-a||<delta ==> ||Fx-Fa||<eps.
Suppose Fix(F) is void and take x_1=0.
Now, if x_1,...,x_n are already defined (n>=1),
we note w_n=||Fx_n-x_n||>0.
Choose x_{n+1} in B such that
||Fx_{n+1}-x_n||<min{w_n/2,delta(x_n,w_n/2)}.
It follows that ||F(F(x_{n+1}))-Fx_n||<w_n/2.
Then we obtain w_n>||F(F(x_{n+1}))-Fx_{n+1}||-w_n=
=3||x_{n+1}-Fx_{n+1}||-w_n.
Consequently, w_{n+1}<=(2/3)w_n, hence w_{n+1}<=(2/3)^{n} w_1 -->0.
On the other hand, ||x_{n+1}-x_n||<w_{n+1}+w_n/2<2(2/3)^{n} w_1.
Thus, (x_n) is Cauchy, hence x_n-->y in B. Since F is continuous and x_n-Fx_n-->0, we get y=Fy, so that Fix(F) is nonempty, a contradiction.

HTH
Best regards,
Ady.
.