Proof of Operations Preserving Convexity of Functions on $\mathbb{R}^n$: Affine Functions, Sums, Nonnegative Multiples, Pointwise Suprema and Affine Substitution

Throughout, $x$ and $y$ are points of the relevant domain, $t\in[0,1]$, and

$$z=(1-t)x+ty$$

(so $z$ lies in the domain, by its convexity); in each part we check the inequality of Convex Function on a Convex Subset of $\mathbb{R}^n$ (i) at $z$.

For (i), bilinearity of the dot product (a background fact of the setting) gives

$$a\cdot z+\beta=(1-t)(a\cdot x+\beta)+t(a\cdot y+\beta),$$

so the convexity inequality holds, even with equality.

For (ii), adding the inequalities $f(z)\le(1-t)f(x)+tf(y)$ and $g(z)\le(1-t)g(x)+tg(y)$ gives

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

For (iii), multiplying the inequality $f(z)\le(1-t)f(x)+tf(y)$ by $\lambda\ge0$ preserves it and gives $\lambda f(z)\le(1-t)\lambda f(x)+t\lambda f(y)$.

For (iv), let $x\in C$. The set $S_{x}=\{f(x):f\in\mathcal{F}\}$ is nonempty, because $\mathcal{F}$ is, and bounded above by hypothesis, so its supremum $F(x)=\sup S_{x}$ exists by The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (ii); by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv) it is an upper bound of $S_{x}$ and lies below every upper bound of $S_{x}$. Now let $x,y\in C$. For every $f\in\mathcal{F}$, convexity of $f$ together with $f(x)\le F(x)$ and $f(y)\le F(y)$ gives

$$f(z)\le(1-t)f(x)+tf(y)\le(1-t)F(x)+tF(y).$$

Thus $(1-t)F(x)+tF(y)$ is an upper bound of $S_{z}$, and therefore $F(z)\le(1-t)F(x)+tF(y)$.

For (v), let $x,y\in C'$. The point $z$ lies in $C'$, since $C'$ is convex by Operations Preserving Convexity in $\mathbb{R}^n$: Intersections, Translates, Scalar Multiples, Sums, Affine Images and Preimages, Closure and Interior (vii) (convexity in $\mathbb{R}^{m}$ as in Segments and Convex Subsets of $\mathbb{R}^n$ (i) and Segments and Convex Subsets of $\mathbb{R}^n$ (ii), with $m$ in place of $n$). By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v), with $m$ and $n$ interchanged, and the vector-space laws of $\mathbb{R}^{n}$,

$$Az+b=(1-t)(Ax+b)+t(Ay+b),$$

where $Ax+b$ and $Ay+b$ lie in $C$. Convexity of $f$ therefore gives

$$f(Az+b)\le(1-t)f(Ax+b)+tf(Ay+b).$$