Lower bound for $\|P^{-1}x\|$

Question

I am working in proving the spectrum of two similar operator are the same. I got in stuck with some steps, which is relating to the norm of inverse operators. More precisely,

Let $H$ be a Hilbert space and $\left \{ f_n \right \}$ be a sequence of unit vectors in $H$.

Assume that $P$ is a bounded invertible operator on $H$.

How can we prove that $\left \| P^{-1}f_n \right \|$ is bounded below by $\frac{1}{\|P\|}$?

I tried as the following.

$\left \| P \right \| = \sup_{\|x\|=1} \|Px\|$. Hence, $\frac{1}{\|P\|} \ge \left \| Pf_n \right \|$.

Moreover, $$\|x\| = \left \| P.P^{-1} x\right \| = \left \| P^{-1} P x \right \| \le \|P^{-1}\| \|Px\|.$$

Hence, $\|P^{-1} \ge \frac{\|x\|}{\|Px\|} = \|P\|$.

It seems to be useless.

Can you give me a hint, please?

Thank you for your time.

Answer 1

Hint: $1=\|f_n\|=\|PP^{-1}f_n\|$

Solution:

We have $1=\|f_n\|=\|PP^{-1}f_n\|\leq\|P\|\|P^{-1}f_n\|$. Multiplying by $\|P\|^{-1}$ we obtain the desired result.