I have a signal $u^*$ and a noisy signal $u_0$. I want to minimize
$$\min_{u\in \mathbb{R}^n} \frac{1}{2}\|u-u_0\|^2+\alpha R(u)$$ where $\alpha>0$ and $R(u) = \|Au\|^2$ with $A=\begin{pmatrix}-1&1&...&...&0\\ 0&-1&1&...&0\\ 0&0&-1&...&0\\ ...&...&...&...&0\\ 0&...&0&-1&1\\ 0&...&0&0&0 \end{pmatrix}$
My attempt was to look at the gradient for every coordinate $u_i$ with $i\in \{1,...n\}$. For example:
$$\frac{d}{du_i} \frac{1}{2}\|u-u_0\|^2+\alpha R(u)\overset{!}{=}0$$ $$\Rightarrow \frac{1}{2}(2u_i-2u_{0_i})+\alpha(4u_i-2u_{i-1}-2u_{i+1}) =0$$ $$u_i = \frac{u_{0_i}+2\alpha u_{i+1}+2\alpha u_{i-1}}{1+4\alpha}$$
As you can see it depends on $u_{i-1}$ and $u_{i+1}$ which I don't know yet. Similarly, $u_1$ depends on $u_2$ and $u_n$ depends on $u_{n-1}$. But I only know the values of the coordinates of $u_0$ so how can I solve it for $u$?
Any suggestions?
You can reduce your expression to a quadratic form: $$ 2 f(\mathbf u) = ||\mathbf u - \mathbf u_0||^2 + 2\alpha ||\mathbf A \mathbf u||^2 = \mathbf u^\prime \left(\mathbf I + 2\alpha \mathbf A^\prime \mathbf A\right) \mathbf u -2 \mathbf u_0^\prime \mathbf u + ||\mathbf u_0||^2 $$ The minimum of this quadratic form is achieved at $$ \mathbf u = \left(\mathbf I + 2\alpha \mathbf A^\prime \mathbf A\right)^{-1} \mathbf u_0 $$