Conditional modulus on a power source ball

If $\delta =0$ and $x$ belongs to the centered-modulus set, then $Kx=0$. By [F3], $x\in N(K{)}^{\perp }$ as well, so $x=0$ and ${\epsilon }_K(X_{\nu ,\rho }(K),0)=0$. If $K=0$, [F3] likewise forces every member of the power source ball to be zero, so the asserted bound holds with no positive singular direction.

Suppose $\delta >0$ and let $x$ be any element of the centered-modulus set. By [F1], choose $w\in X$ with $x=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }w$ and $\mathrm{\parallel }w{\mathrm{\parallel }}_X\le \rho$. Applying [F4] and using $\mathrm{\parallel }Kx{\mathrm{\parallel }}_Y\le \delta$ gives $\mathrm{\parallel }x{\mathrm{\parallel }}_X\le {\delta }^{\nu \mathrm{/}(\nu +1)}{\rho }^{1\mathrm{/}(\nu +1)}$; taking the supremum over the centered-modulus set proves the bound.

Fix a positive singular value $s_j$ and its unit right singular vector $v_j$ from [F5]. Set $w_j=\rho v_j$ and $x_j=\mathrm{\mid }K{\mathrm{\mid }}^{\nu }{w}_j=\rho s_j^{\nu }{v}_j$. Then $\mathrm{\parallel }K{x}_j{\mathrm{\parallel }}_Y=\rho s_j^{\nu +1}={\delta }_j$ and $\mathrm{\parallel }{x}_j{\mathrm{\parallel }}_X=\rho s_j^{\nu }={\delta }_j^{\nu \mathrm{/}(\nu +1)}{\rho }^{1\mathrm{/}(\nu +1)}$. Thus $x_j$ is feasible in the centered-modulus supremum at ${\delta }_j$ and attains the upper bound from step 1.2. If there are infinitely many positive singular values, compactness gives $s_j\to 0$ and hence ${\delta }_j\to 0$; finite positive rank gives only the stated finite list, while $K=0$ was handled in step 1.1.