Is $f$ unitary if $f\in L(V,V)$ such that $a<\frac{\|f^n(v)\|}{\|v\|}<b$, with $0<a<1<b$, for all $n\in \mathbb{N}$ and nonzero $v$

Question

Let $V$ be a complex inner product space. Is $f$ unitary if $f\in L(V,V)$ such that $a<\frac{\|f^n(v)\|}{\|v\|}<b$, with $0<a<1<b$, for all $n\in \mathbb{N}$ and nonzero $v$?

If not, can we say anything about $f^*$ (the adjoint of $f$) and $f$?

Answer 1

No, this inequality is also true if $f=gug^{-1}$ $u$ unitary, $g$ arbitrary. Equivalently $f$ is diagonalisable with eigenvalue of modulus 1. In fact it is an "if and only if". Clearly the hypothesis prove that all eigenvalues are of modulus 1. If it is not diagonalisable, one can find eigenvalue $\lambda$ and two independant vectors with $f(u)=\lambda e, f(v)=\lambda (u+v)$. Then $f^n(v)=\lambda^n(u+nv)$