# $\sum_{n=1}^{\infty}|\left\langle A\phi_{n},\phi_{n}\right\rangle|<\infty\implies A$ is a trace class operator

I'm trying to prove that $\sum_{n=1}^{\infty}|\left\langle A\phi_{n},\phi_{n}\right\rangle|<\infty$ for each orthonormal basis $\{\phi_{n}\}_{n\in\mathbb{N}}$ implies $A$ is in the trace class.

I've used the polar decomposition of $A$ and definition of $Tr(|A|),$ and I've proved that $\sum_{n=1}^{\infty}|\left\langle A\phi_{n},\phi_{n}\right\rangle|^{2}<\infty.$

Then, for the polar decomposition, the expression of $U\phi_{n}=\sum_{n=1}^{\infty}\left\langle U\phi_{n},\phi_{n}\right\rangle\phi_{n}$ and Cauchy-Schwarz inequality we get $$Tr(|A|)=\sum_{n=1}^{\infty}|\left\langle U\phi_{n},A\phi_{n}\right\rangle|\leq(\sum_{n=1}^{\infty}|\left\langle U\phi_{n},\phi_{n}\right\rangle|^{2})^{1/2}(\sum_{n=1}^{\infty}|\left\langle \phi_{n},A\phi_{n}\right\rangle|^{2})^{1/2}$$

So, to inish the proof, we need the first term of the right side of last inequality be finite, but I'm stuck in this.

I've seen two proof in the site but I feel this path could work too, or is there an easier way to prove this?

Also, is there an operator such that $A$ is not in trace class but satisies $\sum_{n=1}^{\infty}|\left\langle A\phi_{n},\phi_{n}\right\rangle|<\infty$ for some orthonormal basis?

Any kind of help is thanked in advanced.

$U\phi_n =\sum_{n=1}^{\infty} \langle U\phi_n , \phi_n \rangle \phi_n$ implies $||U\phi_n||^{2} =\sum_{n=1}^{\infty} |\langle U\phi_n , \phi_n \rangle|^{2}$since $\{\phi_n\}$ is an prthonormal basis. Hence $\sum_{n=1}^{\infty} |\langle U\phi_n , \phi_n \rangle|^{2} < \infty$. For the second question consider the sift operator $T(a_n)=(0,a_1,a_2,...)$ on $l^{2}$. For the standard orthonormal basis $\{e_n\}$ we have $\sum |\langle Te_n,e_n \rangle|=0$ but $T$ is not even compact.
• thanks for answer. The shift operator is exactly the operator that I had in mind. You're totally right about the norm of $U\phi_{n}!$ Thanks a lot! – Squird37 Apr 6 '18 at 5:42