$\min_{x\in\mathbb{R}^n}(\langle z,x-y\rangle+\lVert x-y\rVert^2)=\min_{r\ge0}(-r\lVert z\rVert+r^2)$

Question

Let $y,z\in\mathbb{R}^n$. Then $\min_{x\in\mathbb{R}^n}(\langle z,x-y\rangle+\lVert x-y\rVert^2)=\min_{r\ge0}(-r\lVert z\rVert+r^2)$.

Proof) $-r\lVert z\rVert+r^2$ has the minimum at $r=\frac{\lVert z\rVert}{2}$. Take $x_1$ such that $\lVert x_1-y\rVert=\frac{\lVert z\rVert}{2}$. Then for any $x\in\mathbb{R}^n$, $$ -\lVert x_1-y\rVert\lVert z\rVert+\lVert x_1-y\rVert^2\le -\lVert x-y\rVert\lVert z\rVert+\lVert x-y\rVert^2\le \langle z,x-y\rangle+\lVert x-y\rVert^2.$$ So $\min_{x\in\mathbb{R}^n}(\langle z,x-y\rangle+\lVert x-y\rVert^2)\ge\min_{r\ge0}(-r\lVert z\rVert+r^2)$. How do I prove the converse?

Answer 1

We have $$\min_{x \in \mathbb{R}^n} (\langle z, x-y \rangle + \|x-y\|^2) = \min_{r \ge 0} \min_{x : \|x-y\|=r} (\langle z, x-y\rangle + \|x-y\|^2).$$ The inner minimization problem simplifies to $$\min_{x : \|x-y\|=r} (\langle z, x-y\rangle + \|x-y\|^2) =\min_{x : \|x-y\|=r} \langle z, x-y \rangle + r^2.$$ By Cauchy-Schwarz, $\langle z, x-y \rangle \ge -r\|z\|$, and attains equality when $x = y - r z / \|z\|$.