finite rank operators are dense in trace class operators and Hilbert–Schmidt operators

Question

In many reference books,the proofs were not given,I hope someone can give me hints on how to construct finite rank operators to approximate trace class operators and Hilbert–Schmidt operators.Thnak you in advance!

Answer 1

Rolands answer already gives the idea of how these finite rank operators are constructed. The following argument can be looked up in the proof of Proposition 16.2 / 16.3 from "Introduction to Functional Analysis" by Meise & Vogt (1997).

Any trace-class or Hilbert-Schmidt operator is compact (by construction; in the case of Meise/Vogt) and every compact operator $A\in\mathcal K(\mathcal H)$ on any Hilbert space $\mathcal H$ has a singular value decomposition $A=\sum_{n=1}^\infty s_n(A)\langle \cdot,e_n\rangle f_n$ where $(e_n)_{n\in\mathbb N},(f_n)_{n\in\mathbb N}$ are orthonormal systems in $\mathcal H$ and $(s_n(A))_{n\in\mathbb N}$ is a decreasing null sequence in $[0,\infty)$. Define the sequence of finite-rank operators $A_m:=\sum_{n=1}^m s_n(A)\langle \cdot,e_n\rangle f_n$. From Pythagoras's theorem it follows then that for each $x\in\mathcal H$ and $m\in\mathbb N$

$$ \Vert (A-A_m)x\Vert^2=\Big\Vert \sum_{n=m+1}^\infty s_n(A)\langle x,e_n\rangle f_n\Big\Vert^2\overset{s_n\in[0,\infty)}=\sum_{n=m+1}^\infty s_n(A)^2|\langle x,e_n\rangle|^2 \underbrace{\Vert f_n\Vert^2}_{=1}. $$

Now since $(s_n(A))_{n\in\mathbb N}$ is decreasing, Bessel's inequality yields

$$ \Vert (A-A_m)x\Vert^2=\sum_{n=m+1}^\infty s_n(A)^2|\langle x,e_n\rangle|^2\leq (s_{m+1}(A))^2\sum_{n=m+1}^\infty |\langle x,e_n\rangle|^2\leq (s_{m+1}(A))^2\Vert x\Vert^2. $$

Since $x\in\mathcal H$ was arbitrary, taking the square root on this inequality by the definition of the operator norm gives $\Vert A-A_m\Vert\leq s_{m+1}(A)$ for any $m\in\mathbb N$. The fact that $(s_n(A))_{n\in\mathbb N}$ is null sequence thus shows $\lim_{m\to\infty}\Vert A-A_m\Vert=0$ and concludes this argument.

Answer 2

Not a full answer, but hopefully a nudge in the right direction: Every Hilbert-Schmidt Operator and every trace class operator is a compact operator in a Hilbert space.

In every Hilbert space, every compact operator can be approximated by a series of finite rank operators, this is called the approximation property. So your desired property follows from a more general theorem.

One way I would attempt to prove this, is to take a compact operator $A$, it's spectral decomposition $$A=\sum_{i=1}^\infty= \lambda_i\langle \cdot,e_i\rangle f_i$$ and consider the finite rank operators $$A_n=\sum_{i=1}^n= \lambda_i\langle \cdot,e_i\rangle f_i.$$ I'm pretty sure we can show that $A_n \rightarrow A$ w.r.t. the operator norm.