Consider an unbounded, self-adjoint operator $(T,D(T))$, on a separable Hilbert space $H$. We try to approximate $T$ by its truncation, that is, $T_{n}=P_{n}TP_{n}$, where $P_{n}$ is the projection operator to subspace $span(e_{1},…,e_{n})$. In the strong operator topology(SOT), is $T_{n}$ converges to $T$ strongly? I know that every bounded operator can be approximated by finite rank operator in SOT, does that hold for self-adjoint unbounded operator?
Unbounded self-adjoint operator approximation by bounded operators
functional-analysisoperator-theory
Best Answer
You refer to approximation by bounded operators in the title but in the text you seem to switch to approximation by finite rank operators. Nevertheless, here is a positive answer for the question implied by the title.
Every self-adjoint operator $T$ on a (henceforth assumed separable) Hilbert space is unitarily conjugated to a multiplication operator on $L^2(X)$, where $X$ is a measure space. That is, there is a measurable real function $\varphi $ on $X$ such that $$ D(T) = \{f\in L^2(X): \varphi f\in L^2(X)\} $$ and $T(f)=\varphi f$, for every $f$ in $D(T)$.
For each $n\in {\mathbb N}$, let $X_n=\{x\in X:|\varphi (x)|\leq n\}$, and let $T_n$ be the operator of multiplication by $1_n\varphi $, where $1_n$ denotes the characteristic function of $X_n$. Since $1_n\varphi $ is a bounded function, it is clear that $T_n$ is a bounded operator.
Moreover $T_n (f)\to T(f)$, for every $f$ in $D(T)$. This follows from $$ \|T(f)-T_n(f)\|^2 = \int_X\big |\big (\varphi (x)-1_n(x)\varphi (x)\big )f(x)\big |^2 dx \to 0, $$ thanks to the Lebesgue Dominated Convergence Theorem.