$\| (I-T)^{-1}|_{\ker(I-T)^\perp} \| \geq 1$ for all compact operator $T$ in an infinite dimensional Hilbert space

Question

Let $T$ be a compact operator on a Hilbert space. Prove that

$$\| (I-T)^{-1}|_{\ker(I-T)^\perp} \| \geq 1$$

I'm not even sure how start. Why should $(I-T)^{-1}$ even exist on $\ker(I-T)^\perp$?

Answer 1

A compact operator $T:\mathcal H\to\mathcal H$ cannot have a lower bound on an infinite dimensional subspace $\mathcal M$. Indeed, if there was $c>0$ such that $\|Tx\|\ge c\|x\|$ for all $x\in \mathcal M$, then the image of an orthonormal basis of $\mathcal M$ would be a uniformly separated infinite subset of $\mathcal H$, and therefore not totally bounded.

Rephrase the above as: if $T$ is compact, then for infinite dimensional subspace $\mathcal M$ we have $$\inf\{\|Tx\| : x\in \mathcal M, \ \|x\| = 1\}=0 \tag1$$ hence $$ \inf\{\|I-Tx\| : x\in \mathcal M, \ \|x\| = 1\} \le 1 \tag2$$

Inequality (2) implies that, should the restriction of $I-T$ to $\mathcal M$ be invertible, its inverse will have norm at least $1$.