I've read different references about bounded operators but some are unclear with what they used and others justify but I'm unable to figure out where it comes.
They defined that a linear operator $T: H \to K$, with $H$ and $K$ two Hilbert Spaces, is bounded if $D(T)=H$ ($D(T)$ is the domain of $T$) and $$\sup \{\|Tx\|\ | \|x\|\le 1\} < \infty.$$
The point that bothers me is that $D(T) = H$, some justify it by the fact a bounded linear operator can be extended to $\overline{D(T)}$ and that $D(T)$ is dense in $H$ so we can always work with operator defined everywhere because all the others can be extended to $H$. But why $D(T)$ is dense in $H$ ? Is there the result which states that ? A reference says that most of the operators we use are densely defined, may be they assume that we only work with this kind of operators.
I can solve my problem by thinking that we only use densely or everywhere defined operator but the problem resurfaces when we talk about inverse of a bounded operators. If an operator $T: H \to H$ is 1-1 when can talk about $T^{-1}$ with $D(T^{-1}) = \text{im}(T)$ but I think there is not reason for $T^{-1}$ to be densely or everywhere defined. Then they present the following result : $$T \text{ has a bounded inverse iff } T \text{ is bijective}$$ But what does they mean by inverse ? A everywhere defined inverse ? I have difficulties to prove $\Rightarrow$ because I don't really know what's the definition of bounded inverse.
Here are the references I read :
- Spectral Theory of Operators on Hilbert space, A. D. Andrew and W. L. Green, 2002
- A primary introduction to spectral theory, lecture from Jianwei Yang in 2019.
- Wikipedia and forums.