Maurin (http://www.mscand.dk/article/viewFile/10641/8662) defines nuclear operators like this: A linear operator $A:\mathcal{H}_1\rightarrow \mathcal{H}_2$ where $\mathcal{H}_1$ and $\mathcal{H}_2$ are separable Hilbert spaces is called nuclear if there exist sequences $\{\phi_j\}$ and $\{\psi_j\}$ in $\mathcal{H}_1$ and $\mathcal{H}_2$ respectively such that


for all $\phi\in \mathcal{H}_1$ and $\sum_{j=1}^{\infty}|\phi_j|_{\mathcal{H}_1}|\psi_j|_{\mathcal{H}_2}<\infty$. Now for a trace class operator the polar decomposition


where $\sum_{j=1}^{\infty}\sigma_j(A)<\infty$ shows that $A$ is nuclear. I tried to show every nuclear operator is of trace class but wasn´t successfull so far. Is this true and if why? Many thanks for any help in advance.


1 Answer 1


Assume $A = \sum_j \psi_j \phi_j^*$ be given with $\sum_j \|\psi_j\| \|\phi_j\| < \infty$. Then, $A$ converges in operator norm. In particular $A$ is compact. Thus, $A$ has a singular value decomposition, say $A=\sum_k \sigma_k f_k e_k^*$.

Then, we have $$ 0 \le \sigma_k = (f_k, Ae_k) = \sum_j (f_k, \psi_j)(\phi_j, e_k) $$ and $$ \sum_k \sigma_k = \sum_j \sum_k (f_k, \psi_j)(\phi_j, e_k) = \sum_j \psi_j^* \underbrace{\left(\sum_k f_k e_k^*\right)}_{\|\,.\,\|\le 1} \phi_j \le \sum_j \|\psi_j\| \|\phi_j\| < \infty. $$ That is, $A$ is in the trace class.

  • $\begingroup$ Excellent.Many thanks!! $\endgroup$ Mar 26, 2016 at 12:17
  • $\begingroup$ @user251257 I came across this post and read your answer. I'd be glad if you could explain one detail to me. Why does the inequality between the sums hold? If I use the Cauchy-Schwarz-inequality we still have $\sum_j \sum_k ||f_k||~ ||\psi_j||~ ||\phi_j||~ ||e_k||$. If $f_k$ and $e_k$ are normalized, we're still left with $\sum_j \sum_k ||\psi_j|| ~||\phi_j||$. How do we get rid of the sum over $k$? Or did I miss something obvious? Thank you in advance! $\endgroup$
    – pcalc
    May 10, 2020 at 12:16
  • $\begingroup$ @pcalc you are right. I need to check this. $\endgroup$
    – user251257
    May 10, 2020 at 12:40
  • $\begingroup$ $\sum_k f_k e_k^*$ has operator norm $\le 1$. $\endgroup$
    – user251257
    May 10, 2020 at 12:49
  • $\begingroup$ Would you explain this argument in more detail? $\endgroup$
    – pcalc
    May 10, 2020 at 13:20

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