Hilbert Schmidt operators form a two sided ideal

Vector space. For every finite $F\subseteq E$, [A3] and [A5] give $({\sum }_{e\in F}\mathrm{\parallel }(aS+bT)e{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}\le \mathrm{\mid }a\mathrm{\mid }({\sum }_F\mathrm{\parallel }Se{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}+\mathrm{\mid }b\mathrm{\mid }({\sum }_F\mathrm{\parallel }Te{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }S{\mathrm{\parallel }}_{HS,E}+\mathrm{\mid }b\mathrm{\mid }\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}$, so the finite subsums for $aS+bT$ are bounded by the square of the last quantity [A1] and $aS+bT$ is Hilbert–Schmidt relative to $E$ with the asserted norm bound.

Adjoint. For every Hilbert basis $F$ of $K$, $s_F(T^{\ast })=s_E(T)$ by [A2], so $T^{\ast }$ is Hilbert–Schmidt relative to $F$ and $\mathrm{\parallel }{T}^{\ast }{\mathrm{\parallel }}_{HS,F}=\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}$.

Left multiplication. For finite $F\subseteq E$, [A3] gives ${\sum }_{e\in F}\mathrm{\parallel }ATe{\mathrm{\parallel }}^2\le \mathrm{\parallel }A{\mathrm{\parallel }}^2{\sum }_{e\in F}\mathrm{\parallel }Te{\mathrm{\parallel }}^2\le \mathrm{\parallel }A{\mathrm{\parallel }}^2\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}^2$, so $AT$ is Hilbert–Schmidt relative to $E$ with $\mathrm{\parallel }AT{\mathrm{\parallel }}_{HS,E}\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}$.

Right multiplication. Let $E_0$ and $F$ be the supplied Hilbert bases of $H_0$ and $K$, respectively, and put $U:=TB\in \mathcal{B}(H_0,K)$, so $U^{\ast }=B^{\ast }{T}^{\ast }$ by [A3]. By [step 1.2], $T^{\ast }$ is Hilbert–Schmidt relative to $F$, and [step 1.3] applied to the left multiplication $B^{\ast }{T}^{\ast }$ gives that $U^{\ast }$ is Hilbert–Schmidt relative to $F$ with $\mathrm{\parallel }{U}^{\ast }{\mathrm{\parallel }}_{HS,F}\le \mathrm{\parallel }{B}^{\ast }\mathrm{\parallel }\mathrm{\parallel }{T}^{\ast }{\mathrm{\parallel }}_{HS,F}=\mathrm{\parallel }B\mathrm{\parallel }\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}\mathrm{.}$ Now apply [step 1.2] to the Hilbert–Schmidt operator $U^{\ast }:K\to H_0$, using $F$ as its domain basis and $E_0$ as its codomain basis. It follows that $(U^{\ast }{)}^{\ast }=U^{\ast \ast }$ is Hilbert–Schmidt relative to $E_0$ and has Hilbert–Schmidt norm $\mathrm{\parallel }{U}^{\ast }{\mathrm{\parallel }}_{HS,F}$. Since $U^{\ast \ast }=U$ by [A3], $U=TB$ is Hilbert–Schmidt relative to $E_0$ and $\mathrm{\parallel }TB{\mathrm{\parallel }}_{HS,E_0}=\mathrm{\parallel }{U}^{\ast }{\mathrm{\parallel }}_{HS,F}\le \mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}\mathrm{\parallel }B\mathrm{\parallel }\mathrm{.}$

Both-sided bound. Combining [step 1.3] with $TB$ in place of $T$ and [step 2.1], $ATB=(AT)B$ is Hilbert–Schmidt relative to $E_0$ with $\mathrm{\parallel }ATB{\mathrm{\parallel }}_{HS,E_0}\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }TB{\mathrm{\parallel }}_{HS,E_0}\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}\mathrm{\parallel }B\mathrm{\parallel }$.

Conclusion. Claim 1 is [step 1.1], claim 2 is [step 1.2] and claim 3 is the combination of [step 1.3], [step 2.1] and [step 3.1]; no Hilbert basis is assumed to exist, since $E$, $E_0$ and $F$ are supplied as data and only the finite-subset supremum definition of [A1] and the invariance theorem [A2] are used.