[proofplan] We argue by contradiction. If two distinct points both minimized $f$ on $C$, then convexity of $C$ would place their midpoint back in $C$. Strong convexity gives a strict midpoint inequality at distinct points, forcing the value at the midpoint to be smaller than the common minimum value, which contradicts minimality. [/proofplan]

[step:Assume two distinct minimizers and name their common minimum value] Since $f$ has at least one global minimizer on $C$, choose a point $x_0 \in C$ such that $f(x_0) \le f(x)$ for every $x \in C$. Suppose, toward a contradiction, that there is another global minimizer $y_0 \in C$ with $y_0 \ne x_0$. Define the real number $m \in \mathbb{R}$ by $m := f(x_0)$. Since $y_0$ is also a global minimizer of $f$ on $C$, it attains the same minimum value, so $f(x_0) = m = f(y_0)$. [/step]

[step:Apply strong convexity at the midpoint of the two minimizers]Define the midpoint $z \in \mathbb{R}^n$ by $z := \frac{x_0 + y_0}{2}$. Because $C$ is convex and $x_0, y_0 \in C$, the point $z$ belongs to $C$. By strong convexity of $f$ on $C$, there exists a constant $\alpha > 0$ such that for every $x, y \in C$ and every $t \in [0,1]$, \begin{align*} f((1-t)x + ty) \le (1-t)f(x) + t f(y) - \frac{\alpha}{2}t(1-t)|x-y|^2 \end{align*} Applying this with $x = x_0$, $y = y_0$, and $t = \frac{1}{2}$ gives \begin{align*} f(z) \le \frac{1}{2}f(x_0) + \frac{1}{2}f(y_0) - \frac{\alpha}{8}|x_0-y_0|^2. \end{align*} Using $f(x_0) = f(y_0) = m$, we obtain \begin{align*} f(z) \le m - \frac{\alpha}{8}|x_0-y_0|^2. \end{align*} Since $x_0 \ne y_0$, we have $|x_0-y_0|^2 > 0$, and since $\alpha > 0$, it follows that $f(z) < m$.[/step]

[guided]The midpoint is the natural place to test strong convexity because the definition improves ordinary convexity by a negative quadratic correction whenever the endpoints are distinct. We define \begin{align*} z := \frac{x_0 + y_0}{2}. \end{align*} The convexity of $C$ is needed here: since $x_0, y_0 \in C$ and $\frac{1}{2} \in [0,1]$, the convex combination $\frac{1}{2}x_0 + \frac{1}{2}y_0$ lies in $C$. Thus $z \in C$, so evaluating $f(z)$ is legitimate. By strong convexity, there exists a constant $\alpha > 0$ such that for all $x, y \in C$ and all $t \in [0,1]$, \begin{align*} f((1-t)x + ty) \le (1-t)f(x) + t f(y) - \frac{\alpha}{2}t(1-t)|x-y|^2. \end{align*} We now substitute the specific points $x = x_0$, $y = y_0$, and $t = \frac{1}{2}$. This turns the left-hand side into $f\!\left(\frac{x_0+y_0}{2}\right) = f(z)$, and yields \begin{align*} f(z) \le \frac{1}{2}f(x_0) + \frac{1}{2}f(y_0) - \frac{\alpha}{8}|x_0-y_0|^2. \end{align*} Both $x_0$ and $y_0$ are global minimizers, so they attain the same minimum value $m$: \begin{align*} f(x_0) = m = f(y_0). \end{align*} Substituting these equalities into the midpoint estimate gives \begin{align*} f(z) \le m - \frac{\alpha}{8}|x_0-y_0|^2. \end{align*} The final point is the strictness of the inequality. Since $x_0 \ne y_0$, their Euclidean distance is positive, so $|x_0-y_0|^2 > 0$. Because $\alpha > 0$, the correction term $\frac{\alpha}{8}|x_0-y_0|^2$ is strictly positive, and therefore $f(z) < m$.[/guided]

[step:Derive the contradiction and conclude uniqueness] The point $z$ lies in $C$, while $m$ is the global minimum value of $f$ on $C$. Therefore minimality of $x_0$ gives $m \le f(z)$. This contradicts the inequality $f(z)<m$ obtained above. Hence no two distinct global minimizers of $f$ on $C$ can exist. Since a global minimizer exists by hypothesis, there is exactly one global minimizer of $f$ on $C$. [/step]