Examples of $f$ strictly convex, either with one minimizer or with no minimizer. Let $f\colon X \to [ -\infty, +\infty]$ be proper and strictly convex. Show that $f$ has at most one minimizer. Give examples where $f$ is strictly convex, and either (i) $f$ has one minimizer; or (ii) $f$ has no minimizer.
For the strict convexity part, you may use the fact that f is strictly convex whenever its Hessian is positive definite. 
I am not sure how to show that f has at most one minimizer, could someone help get me started with the question?
 A: Recall the definition of strict convexity:
$$f(\lambda x + (1-\lambda)y) < \lambda f(x) + (1-\lambda) f(y) \quad \forall x,y\in\mathop{\textrm{dom}}(f), ~ x\neq y, ~\lambda\in(0,1)$$
The strictly convex functions $f(x)=e^x$ and $f(x)=x^2$ have exactly zero and one minimizers, respectively.
Now suppose we claim that a strictly convex function $f$ has two or more minimizers. Let $x$ and $y$ be two of those minimizers ($x\neq y$); obviously, $f(x)=f(y)$. Then along the secant between these two points, we have
$$f(\lambda x + (1-\lambda)y) < \lambda f(x) + (1-\lambda) f(y) = \lambda f(x) + (1-\lambda) f(x) = f(x).$$
Hence we have $f(z)<f(x)$, where $z$ is any point strictly between $x$ and $y$---a contradiction. It is not possible therefore for a strictly convex function to have two minimizers (or more).
EDIT: I note that you're using an extended real convention. For the most part, the above argument is unchanged. However, we do need to consider convex functions that obtain $f(x)=-\infty$ at some $x$. Convince yourself of this: that the only convex functions that obtain the value $-\infty$ do so on their entire domain; and the only strictly convex functions that do so have just a single point as their domain.
