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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

$A\phi=\sum_{j=1}^{\infty}(\phi,\phi_j)_{\mathcal{H}_1}\psi_j$

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

$A=\sum_{j=1}^{\infty}\sigma_j(A)(.,\phi_j)\psi_j$

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.

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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 $$ \sum_k \sigma_k = \sum_j \sum_k (f_k, \psi_j)(\phi_j, e_k) \le \sum_j \|\psi_j\| \|\phi_j\| < \infty. $$ That is, $A$ is in the trace class.

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  • $\begingroup$ Excellent.Many thanks!! $\endgroup$ – Peter Melech Mar 26 '16 at 12:17

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