Numerical Inverse Problems and Regularization Error — Examples

Assume Countable Choice. Take $A=\mathrm{diag}(1,{10}^{-3})$ on ${\mathbb{R}}^2$, exact target $x^{†}=0$, noise $e=(0,{10}^{-6})$, observed data $b=e$, and $\lambda ={10}^{-6}$. The regularized inverse sends this noise to a vector of norm $5\cdot {10}^{-4}$, exactly the general bound $\mathrm{\parallel }e\mathrm{\parallel }\mathrm{/}(2\sqrt{\lambda })$.

Assume Countable Choice. In the scalar problem $A=1$, $x^{†}=1$, and $e=0.1$, the data are $b=1.1$. At $\lambda =0.01$, the regularization bias is $1\mathrm{/}101$ and the data-noise contribution has magnitude $10\mathrm{/}101$. At $\lambda =1$, the corresponding magnitudes are $1\mathrm{/}2$ and $1\mathrm{/}20$. This pair of parameter choices shows the tradeoff; it does not assert a universally optimal parameter rule.

Assume Countable Choice. Model a quadrature or collocation approximation by the scalar maps $A=2$ and $A_h=1.9$, exact target $x^{†}=1$, shared data noise $e=0.1$, and $\lambda =1$. Then the exact forward datum is $y=2$, the perturbed datum is $y_{\delta }=2.1$, the discretized perturbed datum is $y_{h,\delta }=2.0$, and the forward residual discarded before inversion is $d_h=0.1$. The refined error chain keeps its propagated residual $R_{\lambda }{d}_h=0.04$ as a separate term.

Assume Countable Choice. Let $A=2$ on $\mathbb{R}$, $\lambda =1\mathrm{/}2$, and $y_{\delta }=1$. The exact regularized solution is $4\mathrm{/}9$. For the approximate value $\tilde{x}=4\mathrm{/}9+1\mathrm{/}100$, the exact normal-equation residual is $9\mathrm{/}200$, so the bound $\mathrm{\mid }\tilde{x}-4\mathrm{/}9\mathrm{\mid }\le \mathrm{\mid }r\mathrm{\mid }\mathrm{/}\lambda$ holds. This uses an exact residual; it does not certify how floating-point residual evaluation error should be bounded.