I am stuck at the following exercise:
Consider a signal $u^*$ and a noisy signal $u_0$. I need to argue that the following problem
$$\min_{u\in \mathbb{R}^n} \frac{1}{2}\|u-u_0\|^2+\alpha S_\mu(u)$$ where $\alpha>0$ and $S_\mu (u) := \sum_{i=1}^n \sqrt{(Au)_i^2+\mu^2} - \mu$ for $A=\begin{pmatrix}-1&1&...&...&0\\ 0&-1&1&...&0\\ 0&0&-1&...&0\\ ...&...&...&...&0\\ 0&...&0&-1&1\\ 0&...&0&0&0 \end{pmatrix}$
has a unique solution.
However, I do not see how to do this. In a similar question this was done by reducing the problem to a quadratic functional, but I do not think that this is possible here due to the presence of roots. Could you please give me a hint?
Define $$\Phi\colon u\mapsto \frac{1}{2}\|u-u_0\|^2 + \alpha S_\mu(u).$$ Notice that $\Phi$ is continuous and $\Phi(u) \to \infty, \|u\|\to\infty$. By Weierstrass theorem, $\Phi$ admits a global minimum $u^*$. Now, note that for each $i \in \{1,\dots,n\}$, the function $u\mapsto \sqrt{(Au)^2_i + \mu^2}$ is convex and the function $u \mapsto \|u-u_0\|^2$ is strictly convex. It follows that $\Phi$ is a strictly convex function (being the sum of convex functions and a strictly convex function). Suppose $v^*\neq u^*$ is another global minimum for $\Phi$. Then, by the strict convexity of $\Phi$, we have \begin{equation*} \Phi\Big(\frac{u^*+v^*}{2}\Big) < \frac{1}{2}\Phi(u^*)+\frac{1}{2}\Phi(v^*) = \Phi(u^*), \end{equation*} a contradition.