Stationary point of $ x \mapsto \frac{1}{2} \| A x - b \|^2 + \frac{\lambda}{2} \| D x \|^2 $

Question

Given full rank matrix $A \in \mathbb{R}^{n \times m}$, matrix $D \in \mathbb{R}^{p \times m}$, vector $b \in \mathbb{R}^n$ and scalar $\lambda > 0$, let scalar field $f: \mathbb{R}^n \to \mathbb{R} $ be defined as follows

$$ f(x)= \frac{1}{2} \| A x - b\|^2 + \frac{\lambda}{2} \| D x \|^2 $$

1. Is $f$ convex?

2. Calculate and classify the stationary point of $f$.

I already prove that $f(x)$ is convex using the fact that $\| \cdot \|^2$ is convex. But I cannot calculate the critical points. Is better to procede using the vector form or using every entry if $f(x)$?

Answer 1

Take the derivative and set it equal to $0$

$$\frac{1}{2}\nabla\left(x^TA^TAx-x^TA^Tb-b^TAx+b^Tb + \lambda x^TD^TDx\right) $$

$$= x^T(A^TA+\lambda D^TD)-b^TA = 0$$

which means

$$x = (A^TA+\lambda D^TD)^{-1}A^Tb$$