I need a really good explication with details of this proof of Hilbert - Schmidt theorem :

Let $(H,\left< , \right>)$ be a complex Hilbert space and let $A:H\rightarrow H$ be a bounded, compact, self-adjoint operator and $(\lambda_n)_n$ a sequence of non-zero real eigenvalues where each eigenvalue of A is repeated in the sequence according to its multiplicity, then there exists an orthonormal set $(v_n)_n$ of corresponding eigenfunctions, i.e. $Av_n=\lambda_nv_n.$ Then A can be written as $$Ax=\sum_n \lambda_n\left<x,v_n\right>v_n,\;\;\;\;\forall\;x\in H. $$

Proof: It's enough to show that $Im A\subset \overline{Sp}(v_n)_n$ using the fact that the closed linear span is dense in H. Because $H=\overline{Sp}(v_n)_n\oplus(\overline{Sp}(v_n)_n)^{\bot}$ and how both subspaces are invariant, it's only remains to show that $$B=A|_{(\overline{Sp}(v_n)_n)^{\bot}}=0.$$ Assume that $B\neq 0$, then there exists a non-zero number $\lambda$ and an eigenvector $v\in(\overline{Sp}(v_n)_n)^{\bot}$ so that $Bv=\lambda v$ and $||v||=1.$ Then $Av=\lambda v$ so $\lambda=\lambda_{n_0},$ for a certain $n_0.$ Therefore $v\in Ker(\lambda_{n_0}I-A)\cap(\overline{Sp}(v_n)_n)^{\bot}=\{0\}$, contradiction.

  • $\begingroup$ Is there a specific line at which you get confused? $\endgroup$ – Mauve Apr 20 '18 at 19:15
  • $\begingroup$ I would say for now the first 4 rows of the proof. $\endgroup$ – Andrew Apr 20 '18 at 19:23

Because $\{v_n\}$ is an orthonormal basis of the closure of its span, if the image of $A$ is contained in this closure then for any $x$ $$ Ax=A\left(\sum_n\langle x,v_n\rangle\,v_n\right)=\sum_n \langle x,v_n\rangle\,Av_n =\sum_n\lambda_n\langle x,v_n\rangle\,v_n $$ (we are using that $\langle Ax,y\rangle=0$ if $y\in\overline{\operatorname{span}}\{v_n:\ n\}^\perp$).

Now, because $A$ is selfadjoint and ${\operatorname{span}}\{v_n:\ n\}$ is invariant for $A$, one can easily show that $\overline{\operatorname{span}}\{v_n:\ n\}^\perp$ is also invariant for $A$.


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