Tikhonov has qualification one for source powers of the Gram operator

On the eigenspace $E_t$ with $t>0$, the exact-data regularization error $x^{†}-x_{\lambda }$ multiplies the source coefficient by $\lambda t^{\nu }\mathrm{/}(t+\lambda )$; on $\ker S$ the source multiplier $t^{\nu }$ is zero. Orthogonality in [F2] and the definition in [F3] therefore give $x^{†}-x_{\lambda }={\sum }_{t\in \mathrm{\Sigma }}\frac{\lambda t^{\nu }}{t+\lambda }{P}_{t}w\mathrm{.}$

Put $s=t\mathrm{/}\lambda$. If $0<s\le 1$, then $s^{\nu }\mathrm{/}(1+s)\le s^{\nu }\le 1$. If $s\ge 1$, then $s^{\nu }\mathrm{/}(1+s)\le s^{\nu -1}\le 1$ because $\nu \le 1$. Thus $\lambda t^{\nu }\mathrm{/}(t+\lambda )={\lambda }^{\nu }{s}^{\nu }\mathrm{/}(1+s)\le {\lambda }^{\nu }$ for every $t>0$.

Applying the orthogonal norm formula to step 1.1 and using step 1.2 gives $\mathrm{\parallel }{x}^{†}-x_{\lambda }{\mathrm{\parallel }}^2={\sum }_{t\in \mathrm{\Sigma }}{\mid \frac{\lambda t^{\nu }}{t+\lambda }\mid }^2\mathrm{\parallel }{P}_{t}w{\mathrm{\parallel }}^2\le {\lambda }^{2\nu }{\sum }_{t\in \mathrm{\Sigma }}\mathrm{\parallel }{P}_{t}w{\mathrm{\parallel }}^2\le {\lambda }^{2\nu }\mathrm{\parallel }w{\mathrm{\parallel }}^2\mathrm{.}$ Taking nonnegative square roots proves the bound. The scalar inequality in step 1.2 is the qualification condition from [F3] with $C_{\nu }=1$.