I'm currently in the process of constructing a proof for the statement below:
If an operator $A$ on $\mathcal{H}$ (Hilbert space) is closed, bounded away from zero, and has a dense range in $\mathcal{H}$, then $A$ has a bounded inverse.
Remarks: $(1)$ An operator $A$ on a Hilbert space $\mathcal{H}$ is a linear map $T:\mathcal{D}(A) \rightarrow \mathcal{H}$, where $\mathcal{D}(A)$ is a dense subspace of $\mathcal{H}$, and $(2)$ An operator $A$ is bounded away from zero if there exists a constant $c>0$ such that $\|Av\| \geq c\|v\|$, for all, $v \in \mathcal{D}(A)$.
I've managed to prove the following claim (which is provided as a hint to prove the main statement):
If an operator $A$ is closed and bounded away from zero, then $A$ is injective and range$(A)$ is closed.
I'm allowed to make use of the following result (known as the Inverse Mapping Theorem):
An operator $A$ has a bounded inverse iff it is closed and bijective.
My question is: if the range$(A)$ is both closed and dense in $\mathcal{H}$, does that imply that $A$ is bijective?
Thanks in advance.
Best Answer
Dense closed subsets coincide with the whole space, this implies that $A$ is onto. Since $A$ is also bounded away from $0$, $A$ is also injective.