# Is $\lambda-R$ left invertible for the right shift operator $R$ and $|\lambda|<1$?

Let $R$ be the right shift operator on $l^{2}(\mathbb{N})$. Then its point spectrum, $\sigma_{p}(R)=\emptyset$. That is, $\lambda- R$ is injective for every $\lambda\in \mathbb{C}$.

If $|\lambda|<1$, is $\lambda- R$ left invertible? For $\lambda=0$, the left inverse is simply the (negative of the) left shift operator.

To prove this, it suffices to show that $\lambda-R$ has closed range or that it is bounded below for $|\lambda|<1.$ I'd be grateful for a hint on how to show this, if true.

By the reverse triangle inequality, $$\|\lambda x - Rx\|\ge \|Rx\|- |\lambda|\|x\| = (1-|\lambda|)\|x\|$$ which shows that $\lambda-R$ is indeed bounded below, hence left-invertible.

The same applies to any norm-preserving operator $T$ (i.e., $\|Tx\|=\|x\|$ for all $x$).