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

Use the value function of variational regularization in its selection-independent formBecause $J\ge 0$, each objective is non-decreasing with $\alpha$, hence so is its infimum. For $0\le t\le 1$, the affine function of the parameter at each fixed $u$ givesThus $\Psi$ is concave.

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