Minimize the following scalar function: $f(x)={\frac 1 2}(x-b)^2+\lambda|x|$ where $b$ and $\lambda \ge 0$ are given real parameters.
I need to show that the function is convex and find the unique minimizer $x^* = x^*(b; \lambda)$.
I figured the second derivative is 1 ($x \ne 0$), showing $f$ is convex. Am I correct in this?
$f' = x - b + \lambda{\frac x {|x|}}$. I am not sure how to solve this for $f' = 0$.
Hint:
The minimum of $f$ on $(-\infty, \infty)$, denoted $$\min_{x\in(-\infty,\infty)} f(x),$$ is equal to $$\min\{\min_{x\in(-\infty, 0]} f(x), \min_{x\in[0,\infty)} f(x)\}$$
Hint: $f'(x)=x-b+\lambda \frac{x}{|x|}$
It means that for $x<0$ $f'(x)=x-b-\lambda$
But for $x>0$ it is $f'(x)=x-b+\lambda$.
Can you finish from here?
$$\frac{1}{2}(x-b)^2+\lambda|x| = \begin{cases} \frac{1}{2}(x-b+ \lambda)^2+\lambda b-\frac{1}{2}\lambda^2 &, x\ge 0\\ \\ \frac{1}{2}(x-b- \lambda)^2-\lambda b-\frac{1}{2}\lambda^2 &, x\lt 0 \end{cases}$$ and to show that this function is convex, refer to kjetil's note.
