# If a linear operator $A$ is closed and injective, then its inverse $A^{-1}$ is also closed

Let $$X$$ and $$Y$$ be Banach spaces and $$A:X\to Y$$ a linear operator. I found the following statement on Wikipedia, https://en.wikipedia.org/wiki/Unbounded_operator#Closed_linear_operators which confuses me.

If $$A$$ is closed (i.e its graph $$\Gamma(A)=\{(x,Ax)):x\in A\}$$ is closed) and injective, then its inverse $$A^{-1}$$ is also closed.

Why is this true? For $$A$$ to have an inverse we need $$A$$ to be surjective as well? Since $$A$$ is closed it is bounded by the closed graph theorem. If $$A$$ would be bijective, then by the inverse mapping theorem, the inverse would be bounded and linear and so closed.

## 1 Answer

Note that this is a statement about unbounded operators. Those are not necessarily defined everywhere. The statement, as a matter of fact, is completely trival, since the graph of $$A$$ is mapped to the graph of $$A^{-1}$$ under the homeomorphism between $$X\oplus Y$$ and $$Y\oplus X$$ which flips the coordinates.

Note also that if $$Dom(A)\subsetneq X$$, then $$A$$ being closed need not imply $$A$$ being bounded. Consider for instance $$X=Y=L^2([a,b]),$$ $$A=\frac{d}{dx}$$ and $$Dom(A)=H^1([a,b])$$. This operator is closed due to completeness of Sobolev Spaces. It is, of course, not very injective.