Let $T_\alpha=A^*A+\alpha I$. For every $f\in X$,
$$
\alpha\|f\|_X^2
\leq\langle f,T_\alpha f\rangle_X
=\|Af\|_Y^2+\alpha\|f\|_X^2
\leq(\|A\|^2+\alpha)\|f\|_X^2.
$$
Thus $T_\alpha$ is a <coercive operator>. If it is invertible and $T_\alpha f=g$, the lower bound and the <Cauchy-Schwarz inequality> give
$$
\alpha\|f\|^2\leq\langle f,g\rangle\leq\|f\|\|g\|,
$$
and therefore $\|T_\alpha^{-1}g\|\leq\alpha^{-1}\|g\|$. Hence
$$
\boxed{\|(A^*A+\alpha I)^{-1}\|\leq\frac1\alpha}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 326 / 2 / b / i / Solution

A bounded operator $T$ on a Hilbert space is coercive when $\operatorname{Re}\langle x,Tx\rangle\geq c\|x\|^2$ for some $c>0$. This estimate implies injectivity and bounds an inverse on the range by $1/c$.


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 Incoming links: Coercive operator