The following is an excerpt in a chapter of Hunter-Nachtergaele's Applied Analysis regarding linear differential operators:
The definition of an unbounded linear operator $A:D(A)\subset H\to H$ acting in a Hilbert space $H$ includes the definition of its domain $D(A)$. We will assume that the domain of $A$ is a dense linear subspace of $H$, unless we state explicitly otherwise. If the domain of $A$ is not dense, then we may obtain a densely defined operator by setting $A$ equal to zero on the orthogonal complement of its domain, so this assumption does not lead to any loss of generality.
Would anyone elaborate the sentence in bold? How does the orthogonal complement of its domain have anything to do with obtaining a densely defined operator?
Best Answer
If $W$ is a linear subspace of a Hilbert space $H$, then $H$ is the orthogonal internal direct sum of the closure of $W$ and the orthogonal complement of $W$, $H=\overline{W}\oplus W^\perp$. If $A:W\to H$ is a linear operator but you wanted to consider a densely defined operator which is "essentially the same" as $A$, using this orthogonal direct sum is the "natural" choice, because $W\oplus W^\perp$ is dense in $H$, and for $x+y\in W\oplus W^\perp$, $x\in W$ and $y\in W^\perp$, you can define $A(x+y)=Ax$.
There are other ways to write $H$ as an internal direct sum $\overline{W}+V$, but those that don't use the orthogonal complement don't preserve the "essentially the same" part by respecting the norm structure of $H$. In particular, if $A$ is a bounded operator, then the extension by setting it to $0$ on $W^\perp$ makes the extension have the same norm. The reason for that is that if $x\in W$ and $y\in W^\perp$, then $\|x+y\|\geq \|x\|$, whereas for an arbitrary internal direct sum $W+V$ you could have $\|x+y\|<\|x\|$, hence the norm can increase for the extension.