Frechet derivative of an homogeneous function

Question

Let be $ f:X\to X$ a homogeneous function of degree 1 definied on a real Banach space $X$ and suppose that exists $u\in X$ such that $f(u)=u$. Let be $L$ the Frechet derivative of $f$ at $u$, i.e. $L=f'(u)$. I would like to prove that $Lu=u$. Any hint?

Answer 1

We have: $$ \lambda u = \lambda f(u) = (1 + \lambda)f(u) - f(u) = f(u + \lambda u) - f(u) = L(\lambda u) + o(\lambda u) = \lambda L(u) + o(\lambda). $$ Now, divide by $\lambda$ and take $\lambda\to 0$.

EDIT: an easy generalization: $f$ homogeneous of degree 1 and $f(u) = v\implies f'(u)u = v$: $$ \lambda v = \lambda f(u) = (1 + \lambda)f(u) - f(u) = f(u + \lambda u) - f(u) = L(\lambda u) + o(\lambda u) = \lambda L(u) + o(\lambda). $$ And link about the o-notation.