Proof of A Local Minimum of a Convex Function is Global

Write $\lVert\,\cdot\,\rVert$ for the Euclidean norm and $|\,\cdot\,|$ for the absolute value; claims 2 and 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ give $d_E(u,v)=\lVert u-v\rVert$ and $\lVert\mu u\rVert=|\mu|\,\lVert u\rVert$. Order arithmetic is taken from Elementary Order Arithmetic in an Ordered Field (claim 2 mixed transitivity, claim 5 product of positive elements, claim 7 inverse of a positive element, claim 8 halving, claim 10 strict compatibility with multiplication by a positive element) and from Elementary Arithmetic in an Ordered Field (claim 3 translation, claim 5 multiplication by a nonnegative element), and the field axioms of the field $\mathbb{R}$ are used for regrouping.

Let $\delta\in\mathbb{R}$ with $0<\delta$ witness the local minimum, so that every $w\in C$ with $d_E(x_0,w)<\delta$ satisfies $f(x_0)\le f(w)$. Let $y\in C$. If $y=x_0$ the conclusion is the reflexivity of $\le$, so assume $y\ne x_0$ and put $\rho=d_E(x_0,y)$. By the axioms of Metric Space we have $0\le\rho$, and $\rho\ne0$ because $y\ne x_0$; hence $0<\rho$.

Choice of the parameter. By halving, $0<\delta/2$ and $\delta/2<\delta$. The element $s=(\delta/2)\,\rho^{-1}$ satisfies $0<s$, since $\rho^{-1}$ is positive and a product of positive elements is positive. Let $t$ be the minimum of $1$ and $s$. By claim 1 of Elementary Properties of the Minimum of Two Elements we have $t\le1$ and $t\le s$, and by claim 2 of that lemma $t$ equals $1$ or $s$, so $0<t$ in either case. Multiplying $t\le s$ by the positive element $\rho$ gives $t\,\rho\le s\,\rho=\delta/2$, and mixed transitivity with $\delta/2<\delta$ gives

Translating $t\le1$ by $-t$ gives $0\le 1-t$.

The comparison point. Put $z=t\,y+(1-t)\,x_0$, which lies in $C$ because $C$ is convex. For each index $i$, distributivity and the field axioms give

the coordinates of sums, scalar multiples and differences of points being those of Sum of Points of $\mathbb{R}^n$, Scalar Multiple of a Point of $\mathbb{R}^n$ and Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$; hence $z-x_0=t\,(y-x_0)$. Therefore, using $|t|=t$ because $0\le t$, and the symmetry axiom of Metric Space,

Conclusion. Since $z\in C$ and $d_E(x_0,z)<\delta$, the local minimum gives $f(x_0)\le f(z)$, and convexity of $f$ gives $f(z)\le t\,f(y)+(1-t)\,f(x_0)$. By transitivity,

Translating by $-\bigl((1-t)\,f(x_0)\bigr)$ and using $f(x_0)-(1-t)\,f(x_0)=\bigl(1-(1-t)\bigr)f(x_0)=t\,f(x_0)$ gives $t\,f(x_0)\le t\,f(y)$. Multiplying by the nonnegative element $t^{-1}$, which is positive because $t$ is, and using $t^{-1}\,(t\,a)=a$ for every $a\in\mathbb{R}$, we obtain $f(x_0)\le f(y)$.