[proofplan] Assume that two distinct points of $C$ both minimize $f$ globally. Convexity of $C$ places their midpoint back in $C$, and strict convexity forces the value of $f$ at that midpoint to be strictly below the average of the two minimal values. Since the two endpoints have the same global minimum value, this produces a point whose value is smaller than the global minimum, a contradiction. [/proofplan]

[step:Assume two distinct global minimizers exist]Suppose, for contradiction, that $x_0,x_1\in C$ are distinct global minimizers of $f$ on $C$. Thus, for every $x\in C$, \begin{align*} f(x_0)\le f(x) \end{align*} and \begin{align*} f(x_1)\le f(x). \end{align*} Applying the [first inequality](/theorems/2897) with $x=x_1$ and the [second inequality](/theorems/2136) with $x=x_0$ gives \begin{align*} f(x_0)\le f(x_1) \end{align*} and \begin{align*} f(x_1)\le f(x_0). \end{align*} Hence \begin{align*} f(x_0)=f(x_1). \end{align*} Define the common minimum value $m\in\mathbb{R}$ by \begin{align*} m=f(x_0)=f(x_1). \end{align*}[/step]

[guided]Suppose, toward a contradiction, that the global minimizer is not unique. Then there exist two different points $x_0,x_1\in C$ which both minimize $f$ on $C$. The phrase “$x_0$ is a global minimizer” means that $f(x_0)$ is no larger than the value of $f$ at any point of $C$: \begin{align*} f(x_0)\le f(x) \end{align*} for every $x\in C$. Likewise, since $x_1$ is also a global minimizer, \begin{align*} f(x_1)\le f(x) \end{align*} for every $x\in C$. Now compare the two minimizing values with each other. Since $x_1\in C$, the global minimality of $x_0$ gives \begin{align*} f(x_0)\le f(x_1). \end{align*} Since $x_0\in C$, the global minimality of $x_1$ gives \begin{align*} f(x_1)\le f(x_0). \end{align*} The two inequalities force equality: \begin{align*} f(x_0)=f(x_1). \end{align*} We denote this common real number by $m$: \begin{align*} m=f(x_0)=f(x_1). \end{align*} This notation records that both alleged minimizers attain exactly the same minimum value.[/guided]

[step:Apply strict convexity at the midpoint] Define the midpoint $z\in V$ by \begin{align*} z=\frac{1}{2}x_0+\frac{1}{2}x_1. \end{align*} Since $C$ is convex and $x_0,x_1\in C$, we have $z\in C$. Since $x_0\ne x_1$ and $\frac{1}{2}\in(0,1)$, strict convexity of $f$ gives \begin{align*} f(z)<\frac{1}{2}f(x_0)+\frac{1}{2}f(x_1). \end{align*} Using $f(x_0)=f(x_1)=m$, this becomes \begin{align*} f(z)<m. \end{align*} [/step]

[step:Contradict global minimality] Because $x_0$ is a global minimizer and $z\in C$, we must have \begin{align*} m=f(x_0)\le f(z). \end{align*} This contradicts the strict inequality \begin{align*} f(z)<m. \end{align*} Therefore two distinct global minimizers cannot exist, so $f$ has at most one global minimizer on $C$. [/step]