# Prove that operator is Bounded below operator

Let $A:X\rightarrow Y$, Where $X,Y$ are Hilbert Spaces, be bounded linear operator. Then show that, for every $\alpha \gt 0$, $A^*A+\alpha I$ is bounded below. Where $A^*=$ Adjoint of $A$.

Attempt: An operator$A$ is called bounded below if there exist $c\gt 0$ such that $$||Ax||\ge c||x||, \forall x\in X$$

$||(A^*A+\alpha I)x||=||A^*Ax+\alpha x||\le(||A^*A||+\alpha)||x||$ I'm stuck here. Help is needed. Thanx in advance.

Hint: Evaluate $$(x, (A^*A+\alpha\,I) \, x)_X$$ and draw the right consequences.
• $(x, (A^*A+\alpha\,I) \, x)=(x,A^*Ax)+(x,\alpha\,x)=||Ax||^2+\alpha ||x||^2\le (||A||^2+\alpha )||x||^2$ what after this? – Chiranjeev_Kumar Feb 3 '17 at 16:59
• No, you have to use $\|A \, x\| \ge 0$. – gerw Feb 3 '17 at 17:55
From $(A^*Ax,x)=(Ax,Ax)=||Ax||^2 \ge 0$, we see that the spectrum of the selfadjoint operator $A^*A$ is contained in $[0, \infty)$. Hence $A^*A+\alpha I$ is invertible ( $\alpha >0$).
Let $B$ be the inverse of $A^*A+\alpha I$. Then for each $x \in X$ we have
$$||x||=||B(A^*A+\alpha I)x|| \le ||B||*|| (A^*A+\alpha I)x||.$$