Hilbert Space & Self Adjoint

Question

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 $A^{-1}:H \rightarrow H$ exists and it is bounded.

My Solution: $A^{-1}:H \rightarrow H$ exists $\Rightarrow$ $A$ is one-to-one then $N(A)={0}$ . Also i know that $N(A)=R(A^{*})^{\bot}$. I don't know how can i proceed solution and first part?!

Answer 1

$\textbf{Hint1:}$ Use the condition given to show that $A$ is injective by showing that $Ax = 0$ implies $x = 0$. Recall that $x = 0$ iff $\|x\| = 0$. This would imply that $A^{-1}$ exists.

$\textbf{Hint2:}$ If $y = Ax$, use the condition given to deduce that $$\|A^{-1}y\| \leq \frac{1}{c}\|y\|.$$ This would show that $A^{-1}$ is bounded.

Answer 2

Also:

• the inequality implies that the image of A is closed.

• since A is self adjoint and injective, the orthogonal complement of the image of A is 0. Iow, the image of A is dense.

These two facts together imply that A is surjective.