2:12 PM
Let $x_0$ be a fixed point of $C$. Then:
$C^2(x_0) = \lambda C (x_0)x_0$
$x_0 = 0 \lor \frac{1}{\lambda}$
and hence $C \neq 0$ thus excluding trivality
Let $C (x_1) = C(x_1 + c)$ for some $c \in \Bbb{C}$. Then:
$C^2 (x_1) = \lambda C (x_1)x_1$
$ C (x_1 + c) = \lambda C (x_1 + c)x_1$
$C (x_1 + c) + \lambda C (x_1 + c) c = \lambda C (x_1 + c) (x_1 + c)$
$C (x_1 + c) + \lambda C (x_1 + c)c = \lambda C^2(x_1 + c)$
$\lambda C^2 (x_1 + c) -C (x_1 + c) (1 + \lambda c) = 0$
$C^2 (x_1+c) = \lambda C (x_1+c)x_1$
$C^2 (x_1+c) + \lambda C (x_1+c) c= \lambda C (x_1+c)(x_1+c)$
$[C^2 (x_1+c) - \lambda C (x_1 + c) (x_1 +c)] = - \lambda C (x_1+c)c$
$\lambda,c=0$ tell us nothing, thus dividing $\lambda, c$ gives
or there exists $x_1$ such that $C(x_1)=0$
meaning $C$ has at least one root
Now, I don't know of any more advanced techniques in nonlinear functional analysis to continue this without flooding the chat thus I will pick it up later
One interesting question, perhaps, is to ask what happens when $C$ is compact
I will be surprised if the literature have few examples of these quadratic looking nonlinear functional equations
