Two fundamental questions about convexity of a function (number1) The first question is as follows (see the second one):

If $f$ is a convex function it is know that there is at most a single minimum. However the argument of the minimum is not guaranteed to be unique. For example $f(x)=c$ is a constant function, therefore it is convex and also concave. Its minimum (also maximum) is $c$. But the whole domain of $f$ is a solution to $\arg \min_x f(x)$. Then the obviious question is: what are the sufficient conditions such that  $\arg \min_x f(x)$ of a convex function $f$ is unique?

My guess is that $f$ must be strictly convex but I dont have any proof.
 A: Strict convexity is by no means the only sufficient condition. Others (below, we use $x^*$ to denote a local minimum):


*

*$f(x)$ is a norm: then by definition, $f(x)\geq 0$ for all $x$, and $f(x)=0$ implies $x=0$. The absolute value function is a simple example here.

*Local strict convexity. That is, $f(x)$ is strictly convex in a small neighborhood $(x^*-\delta,x^*+\delta)$. The Huber penalty function is an example of a function that is locally, but not globally, strictly convex.


You can construct others; for instance, $\|Ax-b\|$, where $A$ has full column rank will have a single local minimum, but it is not strictly convex for all $(A,b)$. (I think it is if $b\not\in\mathop{\textrm{Range}}(A))$.
A: Your intuition is correct.
Definition 0. By the minimizer of a function $f : X \rightarrow \mathbb{R}$, I mean the set of all $x \in X$ such that for all $x' \in X$, we have $f(x) \leq f(x')$.

Proposition 0. Suppose $f : X \rightarrow \mathbb{R}$ is strictly convex, where $X \subseteq \mathbb{R}^n$ is a convex set. Then the
  minimizer of $f$ has at most one element.

Proof. Suppose $x$ and $y$ are elements of the minimizer of $f$. I claim that $x=y$. 
For a contradiction, suppose $x \neq y$.
Idea. Use strict convexity to find a point between $x$ and $y$ at which $f$ is strictly less than $f(x)$.
Execution. Consider any $a,b \in (0,1)$ such that $a+b=1.$ Since the domain of $f$ was assumed convex, we know that $f(ax+by)$ is well-defined. Since $x$ and $y$ are distinct, we know by strict convexity that:
$$f(ax+by) < af(x)+bf(y)$$
Now clearly, $f(x)=f(y),$ since $x$ and $y$ are in the minimizer of $f$. So:
$$f(ax+by) < af(x)+bf(y) = af(x)+bf(x) = (a+b)f(x) = f(x)$$
Hence $f(ax+by)<f(x).$ But this contradicts that $x$ is in the minimizer of $f$, completing the proof.
Addendum. Okay, I was also able to come up with the following:
Definition 1. Suppose $f : X \rightarrow \mathbb{R}$ is a function.


*

*By the local minimizer of $f$, I mean the set of all $x \in X$ such that for some neighbourhood $N$ of $x,$ every $x' \in N$ satisfies $f(x) \leq f(x').$

*By the strict local minimizer of $f$, I mean the set of all $x \in X$ such that for some neighbourhood $N$ of $x,$ every $y \in N \setminus \{x\}$ satisfies $f(x) < f(x').$

Proposition 1. Suppose $f : X \rightarrow \mathbb{R}$ is convex, where $X \subseteq \mathbb{R}^n$ is a convex set. Then the strict local minimizer of $f$ has at most one element.

Proof. Suppose $x$ and $y$ are elements of the strict local minimizer of $f$. I claim that $x=y$.
Suppose for a contradiction that $x \neq y$.
Since $x$ and $y$ are in the strict local minimizer, they're certainly in the (non-strict) local minimizer; hence by convexity, they're in the (global) minimizer. So $f(x)=f(y)$.
Now since $x$ is in the strict local minimizer, there is a neighbourhood $N$ of $x$ such that $f(x) < f(x')$ for all $x' \in N \setminus \{x\}$. Let $r\in \mathbb{R}_{>0}$ satisfy $B_r(x) \subseteq N$. 
We're going to find an element in $B_r(x) \setminus \{x\}$ that is no higher than $x$. Recall that $f(x)=f(y)$. Hence by convexity, for all $a,b \in [0,1]$ satisfying $a+b=1$, we have
$$(*) \qquad f(ax+by) \leq f(x)$$
We now have to play the following game: find $a,b \in [0,1]$ such that $ax+by \in B_r(x) \setminus \{x\}$. Our instincts suggest that we should consider $a=r$ and $b=1-r$. This doesn't quite work, of course; we need to somehow scale by the distance between $x$ and $y$. If $x$ and $y$ are very distant, then $a$ needs to be much smaller. So consider:
$$a = \frac{r}{d(y,x)}, \qquad b = 1-\frac{r}{d(y,x)}$$
Since $r$ is non-zero, hence $a$ is non-zero, so $ax+by$ is indeed distinct from $x$. It remains to show that $ax+by \in B_r(x).$ In other words, we're trying to:
Show: $\|(ax+by)-x\| < r$
We have: $$\|(ax+by)-x\| = \|(a-1)x+by\| = \|\left(\frac{r}{d(y,x)}-1\right) x+\left(1-\frac{r}{d(y,x)}\right)y\|$$
$$=\left|\left(\frac{r}{d(y,x)}-1\right)\right| \cdot \|x-y\| = |r-d(y,x)|<r$$
This basically completes the proof: it follows that $ax+by \in B_r(x) \setminus \{x\}$, hence $ax+by \in N$, hence $f(x)<f(ax+by)$. But by $(*)$, we have $f(ax+by) \leq f(x)$, a contradiction.
