I have this exercise:
Let $T$ a self-adjoint and compact operator in a Hilbert space. Show that there is some compact and self-adjoint operator $S$ such that $S^3=T$.
I have finished this exercise, however the part where I show that $S$ is compact seems longer than expected. I guess that probably will be a shorter way to show the same.
If $T$ is compact and self-adjoint in a Hilbert space $V$ then I know that
$$
Tf=\sum_{k\geqslant 0}\alpha _k \langle f,e_k \rangle e_k,\quad f\in V\tag1
$$
for some orthonormal sequence $(e_n)$ and some real-valued null sequence $(\alpha _n)$. Then its easy to see that setting
$$
Sf:=\sum_{k\geqslant 0}\sqrt[3]{\alpha _k}\langle f,e_k \rangle e_k,\quad f\in V\tag2
$$
then $S^3=T$ and $S$ is self-adjoint by a known result. Now what follows is the part that seems longer than expected:
If we order the sequence $(e_k)$ that defines $T$ such that $|\alpha _k|\geqslant |\alpha _{k+1}|$ for all $k\in \Bbb N $ then its easy to see that $\|T\|=|\alpha _0|$, and so $\|S\|=|\sqrt[3]{\alpha _0}|$. Then define $P_n$ as the orthogonal projection in the space $U_n:=\operatorname{span}(\{e_k:0\leqslant k\leqslant n\})$. Then $(P_nS)$ is a sequence of compact operators, and its immediate to see that $\|S-P_nS\|=|\sqrt[3]{\alpha _{n+1}}|$, and because $(\alpha _n)$ is a null sequence then $(P_nS)\to S$ and so $S$ is compact.
My question: there is a different way than what I did (maybe simpler or shorter) to show that $S$ must be compact?
Best Answer
You need not compute the norms of the operators explicitly due the following:
Lemma
If $(e_n)$ is orthonormal, $a_n \to 0$ and $Sf=\sum a_k\langle f , e_k \rangle e_k$ then $S$ is compact.
Proof: Let $S_Nf=\sum_{k=1}^{N} a_k\langle f , e_k \rangle e_k$. Then $\|Sf-S_Nf\|^{2}= \sum_{k=N+1}^{\infty} |a_k|^{2} |\langle f , e_k \rangle|^{2} \leq b_N \|f\|^{2}$ where $b_N= \sup \{|a_k|^{2}: k>N\}$. Hence the finite rank operators $S_N$ converge to $S$ in operator norm. This makes $S$ compact.