Showing that $(|\left<a, b\right>| - \epsilon)^2 \leq |\left<a, Mb\right>|^2$ for self-adjoint operator $M$ such that $||(I - M)b|| < \epsilon$

Question

Let $a, b$ be $L^2$ normalized functions and $M$ be a self-adjoint pseudodifferential operator such that $||(I - M)b|| < \epsilon$ and $\sigma(M) \leq 1$ (author of the paper I am reading has not defined what $\sigma$ means, so I am assuming it is the norm of $M$).

Edit: Let us furthermore require that $I \neq M \neq 0$.

I am trying to show that $(|\left<a, b\right>| - \epsilon)^2 \leq |\left<a, Mb\right>|^2$, but I seem to be mathematically too immature with operator inequalities. What I have done so far is,

$|\left<a, Mb\right>|^2 = |\left<a, Mb + (I - M)b - (I - M)b\right>|^2 = |\left<a, b - (I - M)b\right>|^2 = |\left<a, b\right> - \left<a, (I - M)b\right>|^2 \geq \left(|\left<a, b\right>| - |\left<a, (I - M)b\right>|\right)^2$

where I need to relate $|\left<a, (I - M)b\right>|$ to $||(I - M)b|| < \epsilon$, but I don't know how.

Any help/tips are appreciated!

Answer 1

By the triangular inequality, Cauchy-Schwarz inequality and the assumptions, you have $$|\langle a,Mb \rangle| \ge |\langle a,b \rangle| - |\langle a,(I-M)b \rangle| \ge |\langle a,b \rangle| - ||a|| \times ||(I-M)b|| \ge |\langle a,b \rangle| - \epsilon.$$ Yet, you cannot square the inequality, since the right-hand side may be negative. You have just $$|\langle a,Mb \rangle| \ge \max \big(|\langle a,b \rangle| - \epsilon,0 \big).$$