I am given the following function for $a^1, a^2, .... , a^n \in \mathbb{R}^n$:
$$f: \mathbb{R}^n \rightarrow \mathbb{R},\, f(x) = \sum_{i=1}^k\left|x-a^{i}\right|_{2}^2$$ where the subscript 2 denotes the euclidean norm.
What am I asked to show is that:
- $f$ has only one critical point $\overline{x}$ (this critical point is the geometric median)
- That $\overline{x}$ is a global minimum
I am stuck with the second part. Here is what I've got so far:
By definition $\overline{x}$ is a critical point if $$\nabla f(\bar{x}) = 0$$
Computing the gradient of $f$:
$$\nabla f(x) = kx - \sum_{i=1}^k a^{i} $$
Setting this equal to zero and solving for $x$:
$$kx - \sum_{i=1}^k a^{i} = 0 \Longrightarrow x = \frac{\sum_{i=1}^k a^{i} }{k} $$
This is the only critical point one can find.
For the second part, I know that $f(\overline{x})$ is a global minimum if $f(x) \geq f(\bar{x})$ for all $x\in \mathbb{R}^n$.
The Hessian Matrix will be a matrix with $k's$ on the diagonal and every other matrix element zero and with $k \in \mathbb{N}$ this means, the Hessian is positive definite which atleast shows that the critical point I have found is a local minimum.
However, being the only local minimum doesn't imply it is the global minimum so how can I show that this critical point is in fact the global minimum?