Spectral interpolation for power source balls

Let $J$ index the positive singular values in [F1]. Write $w=w_0+{\sum }_{j\in J}{w}_j{v}_j$ in norm, with $w_0\in N(K)$. By [F2]–[F3], $x={\sum }_{j\in J}{s}_j^{\nu }{w}_j{v}_j$ and $Kx={\sum }_{j\in J}{s}_j^{\nu +1}{w}_j{u}_j$, with norm-convergent expansions. For each finite $F\subseteq J$, orthonormality gives the squared norms of the partial sums as the sums of squared coefficients; taking limits by continuity of the norm gives $\mathrm{\parallel }x{\mathrm{\parallel }}_X^2={\sum }_{j\in J}{c}_j$ and $\mathrm{\parallel }Kx{\mathrm{\parallel }}_Y^2={\sum }_{j\in J}{a}_j$. The same finite-sum identity and $w_0\perp v_j$ give $\mathrm{\parallel }w{\mathrm{\parallel }}_X^2=\mathrm{\parallel }{w}_0{\mathrm{\parallel }}_X^2+{\sum }_{j\in J}{b}_j\ge {\sum }_{j\in J}{b}_j$, where $c_j=s_j^{2\nu }\mathrm{\mid }{w}_j{\mathrm{\mid }}^2$, $a_j=s_j^{2\nu +2}\mathrm{\mid }{w}_j{\mathrm{\mid }}^2$, and $b_j=\mathrm{\mid }{w}_j{\mathrm{\mid }}^2$.

Put $\theta =\nu \mathrm{/}(\nu +1)\in (0,1)$. For a finite $F\subseteq J$, set $A_F={\sum }_{j\in F}{a}_j$, $B_F={\sum }_{j\in F}{b}_j$, and $C_F={\sum }_{j\in F}{c}_j$. If $A_F=0$ or $B_F=0$, then every $w_j$ with $j\in F$ is zero because $s_j>0$, so $C_F=0$. Otherwise, for $j\in F$ put $p_j=a_j\mathrm{/}{A}_F$ and $q_j=b_j\mathrm{/}{B}_F$. The scalar inequality $t^{\theta }\le \theta t+1-\theta$ holds for $t\ge 0$: for $t>0$ set $g(t)=\theta t+1-\theta -t^{\theta }$. By [F4], $g^{\mathrm{\prime }}(t)=\theta (1-t^{\theta -1})$. If $0<t<1$, strict increase of $u\mapsto u^{1-\theta }$ gives $t^{1-\theta }<1^{1-\theta }=1$; by [F4] and inverse order this implies $t^{\theta -1}>1$ and $g^{\mathrm{\prime }}(t)<0$. If $t>1$, the inequalities reverse and $g^{\mathrm{\prime }}(t)>0$. The mean value theorem then gives $g(t)\ge g(1)=0$ for every $t>0$, while $g(0)=1-\theta >0$. If $q_j>0$, apply the scalar inequality to $t=p_j\mathrm{/}{q}_j$ to obtain $p_j^{\theta }{q}_j^{1-\theta }\le \theta p_j+(1-\theta )q_j$; if $q_j=0$, the same inequality is immediate. Since $c_j=a_j^{\theta }{b}_j^{1-\theta }$, summing and using $\sum p_j=\sum q_j=1$ gives $C_F\le A_F^{\theta }{B}_F^{1-\theta }$.

Exhaust $J$ by its first $n$ indices when $J$ is infinite, and take $F=J$ when it is finite; if $J$ is empty, all three sums in step 1.1 are zero. Passing to the limit in step 1.2 gives $\mathrm{\parallel }x{\mathrm{\parallel }}_X^2\le \mathrm{\parallel }Kx{\mathrm{\parallel }}_Y^{2\theta }\mathrm{\parallel }w{\mathrm{\parallel }}_X^{2(1-\theta )}$. Taking nonnegative square roots and substituting $\theta =\nu \mathrm{/}(\nu +1)$ proves the claimed inequality.