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 $x\in H$. 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 $$ \|Ax\|\geq c\|x\|, \quad\forall x\in H,$$ 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 $(Ax_n)_{n\ge 1}$ be a Cauchy sequence. Then, by the boundedness from below, it holds that $(x_n)_{n\ge 1}$ 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 $\overline{\mathcal{R}(A)}=\mathcal{H}$:$$ \mathcal{R}(A)^{\perp}=\mathcal{N}(A^*)=\mathcal{N}(A)=\{0\} \\ \implies \overline{\mathcal{R}(A)}=\mathcal{R}(A)^{\perp\perp}=\mathcal{H} $$
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.
