Given $f(a,b)=\sum_{k=1}^n(ax_k+b-y_k)^2$ where $x_k$ and $y_k$ are arbitrary real numbers such that $\exists i,k:x_i \ne x_j$.
Show that $f(a,b)$ gets minimized at exactly one point.
I've managed to show that the partial derivatives vanish at exactly one point $(a_0,b_0)$.
I then took the set $K=f^{-1}([f(a_0,b_0),f(a_0,b_0+\varepsilon)]$, and I'm trying to show that $K$ is compact which will finish the proof.
It's clear that $K$ is closed, to show that it is bound I tried to use the fact that $(a,b)\in K\implies |f(a,b)-f(a_0,b_0)| <\varepsilon$ to somehow bound $a-a_0$ and $b-b_0$, but to no avail...
Does anyone know how to bound this set, or maybe have some better idea?