# 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?

• Is $A$ bounded? – Nathanael Skrepek Jul 21 '17 at 10:55
• @NathanaelSkrepek Well... To be honest I do not know. This problem comes from a quantum mechanics book that I am reading and as far as I can tell, no assumption about the boundedness of $A$ is made. – user23823 Jul 21 '17 at 10:58
• see this q&a: math.stackexchange.com/questions/1261307/… – daw Jul 21 '17 at 12:38
• The operator A must be bounded for this to hold. – DanielWainfleet Jul 22 '17 at 1:56
• I have read that if $X$ is a Banach space and $f:X\to X$ is a bounded linear $1$-to-$1$ surjection then $f^{-1}$ is bounded by applying the Uniform Boundedness Principle, but I've never seen how. This principle does not apply if the norm on $X$ is not complete, and I have seen an example where it is incomplete and $f^{-1}$ is unbounded. – DanielWainfleet Jul 22 '17 at 2:49

## 1 Answer

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.

• Would you please explain how your last counterexample (i.e., in the $D\neq\mathcal{H}$ and $A$ unbounded case) can ensure that $Range(A)=\mathcal{H}$? – user23823 Jul 22 '17 at 0:17
• The example $a_i\mapsto\frac1ia_i$ doesn't satisfy the requirement in the question that the range be all of $\mathcal H$. For example, consider the case that $\mathcal H=l^2$ and the $a_i$ are the standard basis vectors (with $i$-th component 1 and all other components 0. – Andreas Blass Jul 22 '17 at 7:05
• More generally, if a bounded linear transformation from one Hilbert space to another is bijective, then its inverse is also bounded, thanks to the closed graph theorem. ("Hilbert" can be weakened to "Banach" here, and probably even further.) – Andreas Blass Jul 22 '17 at 7:07
• @AndreasBlass sorry I messed it up. Hope the edited answer is better. – user60589 Jul 26 '17 at 13:33