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Let $H$ be a Hilbert space and suppose that $A:H rightarrow H$ is a bounded, self-adjoint linear operator such that there is a constant $c>0$ with $c|x| leq |Ax|$ for all $xin H$. Prove that $A^{-1}:H rightarrow H$ exist and it is bounded.

Hint: For ontoness of $A$ it is enough to show that range of $A$ is closed. Why ?

Please help me, if you have any good answer in this question.

Let's see step by step. We say that $A$ is bounded from below if $$
|Ax|geq c|x|, quadforall xin H,$$ holds for some constant $c>0$. We immediately see that $Ax=0$ implies $x=0$, i.e. $A$ is injective. Moreover, we can deduce that $operatorname{ran} A$ is closed. To see this, let $(Ax_n)_{nge 1}$ be a Cauchy sequence. Then, by the boundedness from below, it holds that $(x_n)_{nge 1}$ is Cauchy. Thus, $lim_n Ax_n = A(lim_n x_n) in operatorname{ran} A$. Indeed, $A$ is bounded from below if and only if it is injective and has a closed range. Now, since $A$ is self-adjoint, we know that
$$
operatorname{ran} A = overline{operatorname{ran} }A =ker A^perp =H.
$$ This establishes $A$ is bijective. Let $B$ denote inverse of $A$. Then, by the boundedness from below, we have
$$
|ABx| = |x|geq c|Bx|.
$$ Hence, $B$ is bounded as desired.

Your operator $A$ is clearly injective and closed ranged. For the second property, it is enough so observe that, if $Ax_n to y$, then $|Ax_m - Ax_n| geq c |x_m - x_n|$, hence $(x_n)$ is a Cauchy sequence. By completeness, $x_n to x$, and $Ax = y$.

Since $A$ is self-adjoint, then also $A^*$ is injective and closed ranged. But then $A$ is surjective (for a proof see here).

This proves that $A$ is a bijection. Finally, its inverse operator is continuous by the open mapping theorem.

To show that $overline{mathcal{R}(A)}=mathcal{H}$:$$
mathcal{R}(A)^{perp}=mathcal{N}(A^*)=mathcal{N}(A)={0} \
implies overline{mathcal{R}(A)}=mathcal{R}(A)^{perpperp}=mathcal{H}
$$

Then once you show that the range of $A$ is closed, it follows that $mathcal{R}(A)=mathcal{H}$.

To show that the range is closed, suppose ${ Ax_n }$ converges to $y$. Then ${ Ax_n}$ is a Cauchy sequence, which forces ${ x_n }$ to be a Cauchy sequence because $|Ax_n-Ax_m| ge c|x_n-x_m|$. So ${ x_n }$ converges to some $x$, which gives $Ax=y$. So the range of $A$ is closed.