For $p\in\partial J(v)$, the <Bregman divergence> is
$$
D_J^p(u,v)=J(u)-J(v)-\langle p,u-v\rangle.
$$
Put $r=A\widehat u_{\alpha,\delta}-f^\delta$ and $e=f^\delta-f$, so $\|e\|_Y\leq\delta$. Comparison with $u^\dagger$ gives
$$
\|r\|_Y+\alpha J(\widehat u_{\alpha,\delta})
\leq\delta+\alpha J(u^\dagger).
$$
Using $p^\dagger=A^*w^\dagger$,
$$
\begin{aligned}
D_J^{p^\dagger}(\widehat u_{\alpha,\delta},u^\dagger)
&=J(\widehat u_{\alpha,\delta})-J(u^\dagger)
-\langle w^\dagger,r+e\rangle\\
&\leq
\left(\frac1\alpha+\|w^\dagger\|_{Y^*}\right)\delta
+\left(\|w^\dagger\|_{Y^*}-\frac1\alpha\right)\|r\|_Y.
\end{aligned}
$$
For $0<\alpha<\alpha_0$ the last coefficient is negative, so
$$
\boxed{
D_J^{p^\dagger}(\widehat u_{\alpha,\delta},u^\dagger)
\leq C_\alpha\delta,
\qquad
C_\alpha=\frac1\alpha+\|w^\dagger\|_{Y^*}}.
$$
The estimate holds for any fixed admissible $\alpha$; it does not require $\alpha\to0$ with the noise level.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 326 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-326/1/d/solution