Proof of Affine Functions, Sums, Nonnegative Multiples and Pointwise Suprema of Convex Functions

Fix $x,y\in C$ and $t\in\mathbb{R}$ with $0\le t$ and $t\le1$, and put $z=t\,x+(1-t)\,y$, a point of $C$ because $C$ is convex. Translating $t\le1$ by $-t$ using claim 3 of Elementary Arithmetic in an Ordered Field gives $0\le 1-t$. Claim 5 of that lemma is used for multiplication of an inequality by a nonnegative element, and claim 3 of Elementary Order Arithmetic in an Ordered Field for adding two inequalities; the field axioms of the field $\mathbb{R}$ are used for regrouping, together with the identity $t\,c+(1-t)\,c=\bigl(t+(1-t)\bigr)c=c$ valid for every $c\in\mathbb{R}$.

Claim 1. By claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, applied to the sum and then to each scalar multiple,

$$p\cdot z=p\cdot\bigl(t\,x\bigr)+p\cdot\bigl((1-t)\,y\bigr)=t\,(p\cdot x)+(1-t)\,(p\cdot y).$$

Adding $c=t\,c+(1-t)\,c$ and regrouping gives

$$\ell(z)=t\,\bigl(p\cdot x+c\bigr)+(1-t)\,\bigl(p\cdot y+c\bigr)=t\,\ell(x)+(1-t)\,\ell(y),$$

and an equality is in particular an inequality $\le$, so $\ell$ is convex on $C$.

Claim 2. Convexity of $f$ and of $g$ gives $f(z)\le t\,f(x)+(1-t)\,f(y)$ and $g(z)\le t\,g(x)+(1-t)\,g(y)$. Adding the two inequalities and regrouping with distributivity,

$$(f+g)(z)=f(z)+g(z)\le t\,\bigl(f(x)+g(x)\bigr)+(1-t)\,\bigl(f(y)+g(y)\bigr)=t\,(f+g)(x)+(1-t)\,(f+g)(y).$$

Claim 3. Multiplying $f(z)\le t\,f(x)+(1-t)\,f(y)$ by the nonnegative element $\mu$ and regrouping with distributivity, commutativity and associativity of multiplication,

$$(\mu f)(z)=\mu\,f(z)\le t\,\bigl(\mu\,f(x)\bigr)+(1-t)\,\bigl(\mu\,f(y)\bigr)=t\,(\mu f)(x)+(1-t)\,(\mu f)(y).$$

Claim 4. For each $x\in C$ the set $\{f(x):f\in\mathcal{F}\}$ is nonempty, because $\mathcal{F}$ is nonempty, and is bounded above by hypothesis, so it has a least upper bound by the least upper bound property of The Real Numbers; that least upper bound is unique by Uniqueness of the Supremum and of the Infimum, so $F$ is well defined.

Let $f\in\mathcal{F}$. Since $F(x)$ and $F(y)$ are upper bounds of the respective sets, $f(x)\le F(x)$ and $f(y)\le F(y)$ in the sense of Upper Bound and Least Upper Bound. Multiplying these by the nonnegative elements $t$ and $1-t$ and adding gives

$$t\,f(x)+(1-t)\,f(y)\le t\,F(x)+(1-t)\,F(y),$$

and convexity of $f$ together with transitivity of $\le$ gives $f(z)\le t\,F(x)+(1-t)\,F(y)$. As $f\in\mathcal{F}$ was arbitrary, $t\,F(x)+(1-t)\,F(y)$ is an upper bound of $\{f(z):f\in\mathcal{F}\}$, so the least upper bound $F(z)$ of that set satisfies

$$F(z)\le t\,F(x)+(1-t)\,F(y).$$

Hence $F$ is convex on $C$.