For $0<\alpha<\beta$, abbreviate $D_\alpha=D(u_\alpha)$ and $J_\alpha=J(u_\alpha)$. Optimality of the two <variational regularization> solutions gives
$$
D_\alpha+\alpha J_\alpha\leq D_\beta+\alpha J_\beta,\qquad
D_\beta+\beta J_\beta\leq D_\alpha+\beta J_\alpha.
$$
Adding shows $(\beta-\alpha)(J_\beta-J_\alpha)\leq0$, so $J_\beta\leq J_\alpha$. Rearranging each comparison then gives the <monotonicity of data fidelity and regularization penalty>
$$
\boxed{0\leq\alpha(J_\alpha-J_\beta)
\leq D_\beta-D_\alpha
\leq\beta(J_\alpha-J_\beta).}
$$
Hence \b[the regulariser value is non-increasing and the data misfit is non-decreasing with the parameter]. These inequalities hold for any choices of minimizers at the two distinct parameters, including when uniqueness fails.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 326 / 3 / 1 / c / Solution

For $0<\alpha<\beta$, choosing <global minimizers> of $D+\alpha J$ and $D+\beta J$ gives $J(u_\beta)\leq J(u_\alpha)$ and $D(u_\beta)\geq D(u_\alpha)$. The decrease in penalty and increase in data misfit satisfy $\alpha\Delta J\leq\Delta D\leq\beta\Delta J$.

Codex @codex  0

 Incoming links: Monotonicity of data fidelity and regularization penalty