Power source balls are weighted statistical ellipsoids

Write $w_j=⟨w,v_j{⟩}_X$. By [F4] and the compact spectral multiplier [F5], the coefficient of $\mathrm{\mid }K{\mathrm{\mid }}^{\nu }w$ at $v_j$ is $s_j^{\nu }{w}_j$; the kernel component of $w$ contributes zero.

Conversely, take any $\theta \in {\mathrm{\ell }}^2$ with ${\sum }_j{s}_j^{-2\nu }{\theta }_j^2\le {\rho }^2$ and put $w_j=s_j^{-\nu }{\theta }_j$. This sequence is square-summable with norm at most $\rho$; the coefficient isometry in [F1] gives a unique $w\in N(K{)}^{\perp }$ with these coefficients.

For every $x=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }w$ with $\mathrm{\parallel }w{\mathrm{\parallel }}_X\le \rho$, step 1.1 and Bessel's inequality for the orthonormal family $(v_j)$ give ${\sum }_{j\ge 1}{s}_j^{-2\nu }⟨x,v_j{⟩}_X^2={\sum }_{j\ge 1}{w}_j^2\le \mathrm{\parallel }w{\mathrm{\parallel }}_X^2\le {\rho }^2$. Thus every source element has coefficient sequence in the stated ellipsoid.

Set $x=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }w$ for the vector constructed in step 1.2. By [F5] its $j$th coefficient is $s_j^{\nu }{w}_j={\theta }_j$; [F1] makes equality of the coefficient sequences equality of vectors in $N(K{)}^{\perp }$. Since $\mathrm{\parallel }w{\mathrm{\parallel }}_X\le \rho$, [F2] gives $x\in X_{\nu ,\rho }(K)$. This proves the reverse inclusion.

If $s_{j+1}=s_j$, their inverse powers are equal. Otherwise $0<s_{j+1}<s_j$; by [F7], $f(t)=t^{-\nu }$ is continuous on $[s_{j+1},s_j]$ and differentiable on its interior, with $f^{\mathrm{\prime }}(t)=-\nu t^{-\nu -1}<0$. Thus [F8] gives $f(s_j)-f(s_{j+1})=f^{\mathrm{\prime }}(c)(s_j-s_{j+1})<0$ for some $c\in (s_{j+1},s_j)$. In both cases $s_{j+1}^{-\nu }\ge s_j^{-\nu }$. Hence $a_j=s_j^{-\nu }$ is positive and nondecreasing; by [F3] the right-hand set is a weighted statistical ellipsoid. Steps 2.1 and 2.2 prove the two set inclusions, completing the identification.