Let $E \subset \mathbb R^n$ a closed set, convex and not bounded. Let $f \in C^0 (E, \mathbb R)$ strictly convex, it means that $$\forall (x, y) \in E^2, \forall \alpha \in ]0, 1[, x \neq y \implies f(\alpha x + (1 - \alpha)y) < \alpha f(x) + (1 − \alpha)f(y)$$ such that $$\lim_{\lVert \rVert \to +\infty} f(x) = +\infty ; i.e. ∀ ∈ ]0, +∞[, ∃ ∈ ]0, +∞[, \forall x \in , \lVert x \rVert \geq M \implies f(x) \geq c$$
Show that $f$ has a global minimum and at a unique point.
Since the function is defined on $\mathbb R^n$ I cannot use one dimensional mean value theorem to show that there exists a point $a$ on which $\triangledown f(a)=0$ and I also must show that the hessian matrix is defined positive to show that it is a global minimum but I am running out of ideas to do so.
