Continuously invertible operator.

Question

Let $F$ be a bounded linear operator between the Hilbert spaces $H_1$ and $H_2$. Let $F$ satisfies $$\|z_1-z_2\|\leq \|F(z_1)-F(z_2)\|^t, \ t>0, \ z_1, z_2\in H_1.$$ How to show that $F$ is continuously invertible?

$F$ is one-one is clear. How to show that $F$ is onto?

Answer 1

If $H_1$ is closed subspace of $H_2$ and $F(x)=x$ for all $x \in H_1$ then the inequality holds with $t=1$ but $F$ is not onto. So you cannot prove that $F$ is surjective.

If you assume that $F$ is onto then we can show that $F^{-1}$ is continuous as follows: $F(x_n) \to F(x)$ implies $\|x_n-x\| \leq \|F(x_n)-F(x)\|^{t} \to 0$. This is exactly continuity of $F^{-1}$.

Answer 2

Denote $R=F(H_{1})$, we are to show that $F^{-1}:R\rightarrow H_{1}$ is continuous. Let $y_{n}\rightarrow 0$ be given, then with $z_{n}=F^{-1}(y_{n})$, we have $y_{n}=F(z_{n})$ and $\|z_{n}\|\leq\|F(z_{n})\|^{t}=\|y_{n}\|^{t}\rightarrow 0$, so $z_{n}\rightarrow 0$, in other words, $F^{-1}(y_{n})\rightarrow 0$, so $F^{-1}$ is continuous.