Here’s a proof a standard argument used in analysis, namely the limiting
procedure, also sometimes called a density argument.
Let and be dense in . Suppose that the following inequality holds for some bounded linear operator, , where is a Banach space.
Our goal is to define on .
For all , there exists, by density, a sequence in such that:
Formally we define the limit:
although we have yet to determine if this limit is well-defined. By linearity,
for integers . The left-hand side of converges, so is Cauchy in and is well-defined. Further, the limit is independent of the sequence . Indeed, let denote some other sequence convergence to . Then,