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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.

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Hint: Evaluate $$(x, (A^*A+\alpha\,I) \, x)_X$$ and draw the right consequences.

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  • $\begingroup$ $(x, (A^*A+\alpha\,I) \, x)=(x,A^*Ax)+(x,\alpha\,x)=||Ax||^2+\alpha ||x||^2\le (||A||^2+\alpha )||x||^2$ what after this? $\endgroup$ – Chiranjeev_Kumar Feb 3 '17 at 16:59
  • $\begingroup$ No, you have to use $\|A \, x\| \ge 0$. $\endgroup$ – gerw Feb 3 '17 at 17:55
  • $\begingroup$ you mean to say I should go as Fred already did? can't i get the result by proceeding in my way? $\endgroup$ – Chiranjeev_Kumar Feb 4 '17 at 5:38
  • $\begingroup$ Yes, my hint points in the same direction as Fred's answer, but you do not need to write down the inverse. And no, by your way you only get an upper bound, but you want to have a lower bound. $\endgroup$ – gerw Feb 4 '17 at 9:02
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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

$$||x||=||B(A^*A+\alpha I)x|| \le ||B||*|| (A^*A+\alpha I)x||.$$

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