Invertibility of one-to-one and onto linear operators on a Hilbert space $\mathcal{H}$ is a Hilbert space. $D$ is some dense subspace of $\mathcal{H}$. A linear operator $A:D\rightarrow \mathcal{H}$ is one-to-one (i.e., $Au=0$ implies $u=0$) and onto (i.e., $Range(A)=\mathcal{H}$). Do these conditions suffice to establish that $A$ has a bounded inverse? 
 A: I will treat two cases: First $A$ is bounded and then $A$ is unbounded.
Assume $A$ is bounded and let $x \notin D\neq \mathcal{H}$, then, since $D$ is dense, there is a Cauchy sequence $x_i \in D$ such that 
$$\lim x_i = x $$
because $A$ is bounded $A(x_i)$ is a Cauchy sequence, hence convergent. Anticipatory I will denote its limit by $A(x)$.
If $A$ would have a bounded inverse then
$$ A^{-1}(A(x) )= \lim  A^{-1}(A(x_i) )=\lim x_i = x = x$$
Hence $x\in Range(A^{-1})$ which is a contradiction. So when $A$ is bounded it is nessary to have $D= \mathcal{H}$ and in this case Banach-Schauer-Theorem applies which tells you $A^{-1}$ is always bounded. 
The second case: Do we have an unbounded linear isomorphism? Just take a linear independent convergent sequence $\{x_i\}$ such that the limit $x$ is also linear independent with $\{x_i\}$. (For instance you could take the series expression of the exponential function in $L^2$) Then map $x_i$ to $x_i$ and $x$ to $2x$. And extend it to a linear isomorphism. (This can be done by linear algebra if $D$ and $\mathcal{H}$ have the same cardinality) But this map can't be continuous, hence it is not bounded. Now we can take $A^{-1}$ to be such an map.
Counter-exmaples to Banach-Schauder-Theorem in this case are also interesting.
So what is left is the question for which unbounded operators $A$ the inverse is bounded. But if you look at the inverse $A^{-1}$ you want an injective operator with dense image. For normal operators we have that being injective is equivalent to having dense image. So you might want to check if $A$ being normal implies that $A^{-1}$ is normal. 
