Proof of Jensen's Inequality for Finite Convex Combinations in $\mathbb{R}^n$

We prove by induction on $N\in\mathbb{N}$ that the inequality of (i) holds for all $x_{1},\dots,x_{N}\in C$ and all $t\in\Delta_{N}$. Throughout, a sum over $[N+1]$ splits as the sum over $[N]$ plus its last term, and a one-term sum is its term, by Iterated Operations: Finite Sums and Finite Products (i) and Iterating a Binary Operation along a Finite List (i).

For $N=1$, Convex Combinations: Small Cases, Splitting Off the Last Point, Dropping a Zero Weight, Joining Two Combinations, and Membership in Convex Sets (i) gives $t_{1}=1$ and $\sum_{k=1}^{1}t_{k}x_{k}=x_{1}$, while $\sum_{k=1}^{1}t_{k}f(x_{k})=f(x_{1})$; so the inequality holds with equality.

Now assume the claim for some $N$, and let $x_{1},\dots,x_{N+1}\in C$ and $t\in\Delta_{N+1}$. Put

$$z=\sum_{k=1}^{N+1}t_{k}x_{k}.$$

By Convex Combinations: Small Cases, Splitting Off the Last Point, Dropping a Zero Weight, Joining Two Combinations, and Membership in Convex Sets (iii), $0\le t_{N+1}\le1$, and we treat the cases $t_{N+1}=1$ and $t_{N+1}<1$ separately.

If $t_{N+1}=1$, then $z=x_{N+1}$ by Convex Combinations: Small Cases, Splitting Off the Last Point, Dropping a Zero Weight, Joining Two Combinations, and Membership in Convex Sets (iii). Since $\sum_{k=1}^{N+1}t_{k}=1$ (Convex Weights, the Standard Simplex, and Convex Combinations in $\mathbb{R}^n$ (i)), the sum $\sum_{k=1}^{N}t_{k}$ is $0$, and each of its nonnegative terms lies between $0$ and this sum by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (vi). Hence $t_{k}=0$ for every $k\in[N]$, and $\sum_{k=1}^{N}t_{k}f(x_{k})=\sum_{k=1}^{N}0=0$ by Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (iv). Therefore

$$\sum_{k=1}^{N+1}t_{k}f(x_{k})=f(x_{N+1})=f(z).$$

If $t_{N+1}<1$, then Convex Combinations: Small Cases, Splitting Off the Last Point, Dropping a Zero Weight, Joining Two Combinations, and Membership in Convex Sets (iii) gives $s=1-t_{N+1}>0$ and convex weights $\tau=(t_{1}/s,\dots,t_{N}/s)\in\Delta_{N}$ such that, with $y=\sum_{k=1}^{N}\tau_{k}x_{k}$,

$$z=(1-t_{N+1})\,y+t_{N+1}\,x_{N+1}.$$

The point $y$ lies in $C$ by Convex Combinations: Small Cases, Splitting Off the Last Point, Dropping a Zero Weight, Joining Two Combinations, and Membership in Convex Sets (vi). We now apply, in turn, the convexity inequality of Convex Function on a Convex Subset of $\mathbb{R}^n$ (i), the induction hypothesis for $x_{1},\dots,x_{N}$ and $\tau$ (multiplied by $s>0$), and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (i) with $s\tau_{k}=t_{k}$:

$$f(z)\le s\,f(y)+t_{N+1}f(x_{N+1})\le s\sum_{k=1}^{N}\tau_{k}f(x_{k})+t_{N+1}f(x_{N+1})=\sum_{k=1}^{N+1}t_{k}f(x_{k}).$$

This completes the induction.