Moore--Penrose inverse of a Hilbert-space operator

Let $A:X\to Y$ be a bounded operator between Hilbert spaces. Write $N=\ker A$ and $R=\mathrm{im}A$. The possibly unbounded Moore-Penrose inverse has domain of a functionFor $y=Ax+z$ with $z\in R^{\perp }$, define $A^{†}y$ to be the unique vector in $N^{\perp }$ mapped by $A$ to $Ax$. Equivalently,Here $P_{N^{\perp }}$ is an orthogonal projection. The definition is independent of the chosen $x$, since two preimages differ by an element of the kernel. It also specifies the action on $R^{\perp }$: that component is sent to zero.

This inverse is a bounded operator exactly when $R$ is a closed set. If $R$ is closed, the restriction $A:N^{\perp }\to R$ is a bounded bijection of Hilbert spaces, and the bounded inverse theorem applies. Conversely, a bound $\Vert A^{†}y\Vert \le C\Vert y\Vert$ implies $\Vert x\Vert \le C\Vert Ax\Vert$ for $x\in N^{\perp }$. A convergent sequence $(A{x}_j)$ then gives a Cauchy sequence $(x_j)$ in $N^{\perp }$; taking its limit proves that $R$ is closed.

For an infinite-dimensional counterexample to automatic boundedness, consider $A{e}_n=e_n/n$ on the l2 sequence space. The inverse sends $e_n$ to $n{e}_n$ and is unbounded. Its domain is therefore important: unlike the finite-dimensional matrix case, a Moore-Penrose inverse need not be an everywhere-defined bounded map.