= Solution
A function $f:C\to\mathbb R$ is <strictly convex> when, for distinct $x,y\in C$ and $0<t<1$,
$$
\boxed{f(tx+(1-t)y)<tf(x)+(1-t)f(y)}.
$$
Suppose distinct points $x$ and $y$ were both minimizers, with minimum value $m$. Since $C$ is a <convex set>, their midpoint belongs to $C$, while strict convexity gives
$$
f\left(\frac{x+y}{2}\right)
<\frac{f(x)+f(y)}2=m,
$$
a contradiction. Thus any minimizer is unique, which is the <uniqueness of a minimizer of a strictly convex function>.
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