= Solution
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
$$
\boxed{\|(A^*A+\alpha I)^{-1}\|\leq\frac1\alpha}.
$$
Back to article page
