Arguing that local minimum is global minimum

Question

Let $$f(x)=\sum_{v=1}^k m_v\|x-x_v\|_2^2, \ \ \ x,x_v \in \mathbb{R}^n, m_v \in \mathbb{R}>0$$ Then $$\nabla f(x) = 2 \sum_{v=1}^k m_v(x-x_v) = 0$$ So $f$ has the local minimum $$x = \frac{\sum_{v=1}^k m_vx_v}{\sum_{v=1}^k m_v}$$

How can I argue that this is also a global minimum?

Answer 1

HINT:

substitute the values of $x$ in $f(x)$ and check the value for every x (after doing the second derivative test).the largest one is the critical point and the smallest one is the stable point.