Use the <value function of variational regularization> in its selection-independent form
$$
\Psi(\alpha)=\inf_u\{D(u)+\alpha J(u)\},\qquad 0\leq\Psi(\alpha)\leq M:=D(0).
$$
Because $J\geq0$, each objective is non-decreasing with $\alpha$, hence so is its <infimum>. For $0\leq t\leq1$, the <affine function> of the parameter at each fixed $u$ gives
$$
\begin{aligned}
\Psi(t\alpha+(1-t)\beta)
&=\inf_u\{t\Phi_{\alpha,f}(u)+(1-t)\Phi_{\beta,f}(u)\}\\
&\geq t\Psi(\alpha)+(1-t)\Psi(\beta).
\end{aligned}
$$
Thus $\Psi$ is <concave>.

For continuity, on any interval $0<a\leq\alpha<\beta\leq b$, comparison using $u_\alpha$ gives
$$
0\leq\Psi(\beta)-\Psi(\alpha)
\leq(\beta-\alpha)J(u_\alpha)
\leq\frac{M}{a}(\beta-\alpha).
$$
It follows that $\Psi$ is a <locally Lipschitz function>, hence <continuous>, on $(0,\infty)$. Therefore \b[the <value function of variational regularization> is non-decreasing, <concave> and <continuous>]. No uniqueness of minimizers is needed for these conclusions.

Solution

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

{title2=$\Psi(\alpha)=\inf_u[D(u)+\alpha J(u)]$}

For a nonnegative regulariser $J$, the optimal objective value is non-decreasing and <concave> in the positive regularization parameter. If $J(0)=0$ and $D(0)<\infty$, comparison with zero makes it finite and a <locally Lipschitz function> away from parameter zero.

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