Problem: Given the operator $ \phi : L \to L $ in an inner-product space $L$ over $\mathbb{C} $. Prove: If there exists $ v \in L $ s.t. $ \left\lVert \phi(v) \right\rVert < \left\lVert \phi^*(v) \right\rVert $, then there exists $ u \in L $ s.t. $ \left\lVert \phi(u) \right\rVert > \left\lVert \phi^*(u) \right\rVert $
I don't know how to approach the solution. I don't have any other information about $ \phi $ ( maybe the question's lacking more information, it appeared in my course's lecture notes few years ago ). I thought maybe I'd look at $ \psi = \phi \phi^* + \phi^* \phi $ which is hermitian, then since $ \psi $ is hermitian, there exists orthonormal basis $ B = \{ v_1 ,...,v_n \} $ of $ V $ for which $ \psi $ is diagonal.
Then, assuming there exists $ v \in L $ s.t. $ \left\lVert \phi(v) \right\rVert < \left\lVert \phi^*(v) \right\rVert $, we can write $ v = \sum_{i=1}^n \alpha_i \cdot v_i $ ( $ v $ is spanned by the basis vectors ), and so $ \left\lVert \phi(v) \right\rVert < \left\lVert \phi^*(v) \right\rVert $ $ \iff $ $ \langle \phi(v),\phi(v) \rangle < \langle \phi^*(v),\phi^*(v) \rangle $ $\iff $ $ \sum_{i=1}^n \alpha_i \langle \phi(v_i),\phi(v) \rangle < \sum_{i=1}^n \bar{\alpha_i} \langle \phi^*(v_i),\phi^*(v) \rangle $
maybe this is the direction I should proceed somehow ( I haven't even used $ \psi $ ).
I'm lost, can you please help as to how to prove the theorem?
Thanks in advance for any help!
This is false. On $\ell^{2}$ let $\phi ((x_n))=(x_2,x_3,x_4,..)$. Then $\phi^{*}((x_n))=(0,x_1,x_2,...)$. If $x=(1,0,0,...)$ then $\|\phi (x)\|=0<1=\|\phi^{*}(x)\|$. But there is no $x$ such that $\|\phi^{*}(x)\| <\|\phi (x)\|$.