Let $d\in\mathbb{R}$ and $\alpha>0$ be given.
(i) Explain why the function $f(x)=\dfrac{1}{2}(x-d)^2+\alpha|x|$ is strictly convex.
(ii) verify that \begin{equation} \bar{x}=(|d|-\alpha)_{+}\;\text{sign(d)}=\begin{cases} d+\alpha,\qquad\text{if }\;\;\;\;\;\;d<-\alpha\\ 0,\qquad\;\;\;\;\;\;\text{if}\;\; -\alpha\leq d\leq \alpha\\ d-\alpha\qquad\;\;\text{if}\;\;\;\;\;\;\alpha<d \end{cases} \end{equation} is a minimizer of $f$ and therefore $\bar{x}$ is unique. (I believe $|d|-\alpha)_{+}\;\text{sign(d)}$ means we only care about positive values)
I am looking for confirmation that I am on the right path with this problem.
For part (i) I think it should follow as such:
- Show $f$ has a minimum via differential calculus. That is, set $f'(x)=0$ and find the critical point.
- Show that this minimum is unique
- Show that $0\in\partial f(\bar{x})$ (i.e. Generalized Fermat theorem) which will give strict convexity
For part (ii), intuitively I want to simply substitute each case value into $f'(\bar{x})=0$, and show that it holds; however, it only seems to hold for the case where $d<-\alpha$.