The Hilbert–Schmidt norm is basis independent

For every $e\in E$ the vector $Te$ lies in $K$, so [F1] applied in $K$ to the Hilbert basis $F$ gives $\mathrm{\parallel }Te{\mathrm{\parallel }}^2={\sum }_{f\in F}{a}_{e,f}$, a supremum over finite $B\subseteq F$.

For every $f\in F$ the vector $T^{\ast }f$ lies in $H$ by [F2], so [F1] applied in $H$ to the Hilbert basis $E$ gives $\mathrm{\parallel }{T}^{\ast }f{\mathrm{\parallel }}^2={\sum }_{e\in E}\mathrm{\mid }⟨T^{\ast }f,e⟩{\mathrm{\mid }}^2$; since $⟨T^{\ast }f,e⟩=\overline{⟨e,T^{\ast }f⟩}$ and $⟨Te,f⟩=⟨e,T^{\ast }f⟩$ by [F2], the moduli agree: $\mathrm{\mid }⟨T^{\ast }f,e⟩\mathrm{\mid }=\mathrm{\mid }⟨Te,f⟩\mathrm{\mid }=a_{e,f}^{1\mathrm{/}2}$.

The iterated suprema agree with the rectangle supremum. For every finite $A\subseteq E$ the identity $\sup \{{\sum }_{e\in A}{\sum }_{f\in B}{a}_{e,f}:B\subseteq F\text{ finite}\}={\sum }_{e\in A}{\sum }_{f\in F}{a}_{e,f}$ holds. If $A=\mathrm{\varnothing }$, both sides are zero; hence assume $A\mathrm{\ne }\mathrm{\varnothing }$. Each row sum is the finite number $\mathrm{\parallel }Te{\mathrm{\parallel }}^2$ by step 1.1. The left side is at most the right side because each $B$ gives a subsum, while for the reverse inequality fix a real $\eta >0$ and, using [F3], choose for each $e\in A$ a finite $B_e\subseteq F$ with ${\sum }_{f\in B_e}{a}_{e,f}>{\sum }_{f\in F}{a}_{e,f}-\eta$; then $B:={\bigcup }_{e\in A}{B}_e$ is finite and ${\sum }_{e\in A}{\sum }_{f\in B}{a}_{e,f}\ge {\sum }_{e\in A}{\sum }_{f\in B_e}{a}_{e,f}>{\sum }_{e\in A}{\sum }_{f\in F}{a}_{e,f}-\mathrm{\mid }A\mathrm{\mid }\eta$. Hence the supremum over all finite rectangles $A\times B$ equals ${\sup }_A{\sum }_{e\in A}{\sum }_{f\in F}{a}_{e,f}={\sup }_A{\sum }_{e\in A}\mathrm{\parallel }Te{\mathrm{\parallel }}^2=s_E(T)$ by [step 1.1] and [F4]; and since every finite $S\subseteq E\times F$ is contained in a rectangle while subsums are monotone, this rectangle supremum is also the supremum over all finite subsets of $E\times F$.

The same computation with the adjoint. By [step 1.2] and the same argument with $E$ and $F$ interchanged, $s_F(T^{\ast })={\sup }_B{\sum }_{f\in B}{\sum }_{e\in E}\mathrm{\mid }⟨T^{\ast }f,e⟩{\mathrm{\mid }}^2={\sup }_{A,B}{\sum }_{e\in A}{\sum }_{f\in B}\mathrm{\mid }⟨T^{\ast }f,e⟩{\mathrm{\mid }}^2={\sup }_{A,B}{\sum }_{e\in A}{\sum }_{f\in B}{a}_{e,f}$, the last equality by the modulus identity of [step 1.2]; the middle supremum is over finite rectangles, and it is the finite-subset supremum of $E\times F$ because finite subsets of a product lie in rectangles.

Conclusion of the matrix-coefficient form. Steps 2.1 and 2.2 identify the rectangle supremum of claim 1 with $s_E(T)$ and with $s_F(T^{\ast })$ respectively, so that supremum equals both sums; this proves claim 1.

Basis independence. Let $E^{\mathrm{\prime }}$ be any Hilbert basis of $H$. Applying [step 3.1] to the pair $(E^{\mathrm{\prime }},F)$ gives $s_{E^{\mathrm{\prime }}}(T)=s_F(T^{\ast })$, and applying it to $(E,F)$ gives $s_E(T)=s_F(T^{\ast })$ for the same basis $F$ of $K$; hence $s_E(T)=s_{E^{\mathrm{\prime }}}(T)$, and also $s_E(T)=s_F(T^{\ast })$ for every Hilbert basis $F$ of $K$, both equalities holding in $[0,+\mathrm{\infty }]$.

Membership and the norms. By [step 4.1] the sums $s_E(T)$, $s_{E^{\mathrm{\prime }}}(T)$ and $s_F(T^{\ast })$ all equal one extended real number, so they are finite simultaneously; when the common value is finite, taking nonnegative square roots gives $\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E}=\mathrm{\parallel }T{\mathrm{\parallel }}_{HS,E^{\mathrm{\prime }}}=\mathrm{\parallel }{T}^{\ast }{\mathrm{\parallel }}_{HS,F}$ by [F6], and when it is $+\mathrm{\infty }$ none of the three norms is defined and $T$ is Hilbert–Schmidt relative to no Hilbert basis of $H$. This is claim 3.