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.
Best 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.