Convexity

Put $\lambda=(z-y)/(z-x)\in(0,1)$; then $y=\lambda x+(1-\lambda)z$, so convexity gives $f(y)\le\lambda f(x)+(1-\lambda)f(z)$.

Multiplying this inequality by the positive number $z-x$ and rearranging gives $(z-y)(f(y)-f(x))\le(y-x)(f(z)-f(y))$.

Dividing step 2.1 successively by the positive products $(y-x)(z-x)$ and $(z-y)(z-x)$ gives $s(x,y)\le s(x,z)\le s(y,z)$.

Choose $u,v\in I$ with $u<a\le b<v$; then for $a\le x<y\le b$, two applications of the three-slope inequality give $s(u,a)\le s(x,y)\le s(b,v)$.

With $M:=\max\{|s(u,a)|,|s(b,v)|\}$, step 1.1 yields $|f(y)-f(x)|\le M|y-x|$ for $x<y$; symmetry gives the same estimate for all $x,y\in[a,b]$, which is the Lipschitz condition.

Given $c\in\operatorname{int}I$, choose such an interval $[a,b]$ containing $c$ in its interior; the estimate in step 2.1 gives the $\varepsilon$--$\delta$ condition at $c$ by taking $\delta=\varepsilon/M$ when $M>0$, and is immediate when $M=0$.

For fixed $c\in I$, the functions $x\mapsto s(x,c)$ on $x<c$ and $x\mapsto s(c,x)$ on $x>c$ are nondecreasing by [L1]; choosing points on both sides of $c$, [L1] bounds each near $c$ between two fixed finite outer secant slopes.

The monotone one-sided-limit theorem [L3] therefore supplies finite one-sided limits of these two slope functions at $c$, and [L2] identifies them respectively with $f'_-(c)$ and $f'_+(c)$.

Apply [L1] to $x<u<v$ and let $x\to u^-$, then to $u<v<z$ and let $z\to v^+$; together with $s(u,v)\le f'_-(v)$ and $f'_+(u)\le s(u,v)$ obtained in the same way, this gives the displayed chain.

Put $g(c):=f'_+(c)$. The order chain in [L1] gives $g(u)\le g(v)$ whenever $u<v$, so $g$ is nondecreasing.

For $u<c$, [L1] gives $g(u)\le f'_-(c)$. Conversely, if $a<u<c$, [L1] gives $s(a,u)\le f'_-(u)\le g(u)$. Letting $u\to c^-$, continuity from [L3] gives $\lim_{u\to c^-}s(a,u)=s(a,c)$; then letting $a\to c^-$ gives $\lim_{u\to c^-}g(u)=f'_-(c)$. Since $g(c)=f'_+(c)$, nondifferentiability of $f$ at $c$ makes $g$ discontinuous there.

Froda's theorem makes the discontinuity set of $g$ at most countable, and step 1.2 places every nondifferentiability point of $f$ in that set.

If $x<c$, [L1] applied to $x<c$ gives $(f(c)-f(x))/(c-x)\le f'_-(c)\le m$.

If $x>c$, [L1] gives $m\le f'_+(c)\le(f(x)-f(c))/(x-c)$.

Multiplying the inequalities in steps 1.1 and 2.1 by their positive denominators and rearranging yields $f(x)\ge f(c)+m(x-c)$ on both sides of $c$, while equality holds at $c$; hence [L2] applies.

Assume $f$ is convex. For $u<v$, [L1] becomes $f'(u)\le f'(v)$, so $f'$ is nondecreasing.

Assume instead that $f'$ is nondecreasing. For a fixed $c$, the derivative of $g(x)=f(x)-f'(c)x$ is nonpositive on the left of $c$ and nonnegative on the right; apply [L2] to $-g$ on the left and to $g$ on the right to obtain $f(x)\ge f(c)+f'(c)(x-c)$ for every $x\in I$.

If $z=\lambda x+(1-\lambda)y$, multiply the supporting inequalities of step 1.2 at $z$ for $x$ and $y$ by $\lambda$ and $1-\lambda$ and add; this gives $f(z)\le\lambda f(x)+(1-\lambda)f(y)$. Thus the two cases prove the equivalence.

By [L1], $f'$ is nondecreasing on $I$.

Thus $f'$ is monotone, so [L2] gives continuity of $f'$.

The conclusion holds at every point of the open interval $I$.

Assume $f$ is convex. By [L1], $f'$ is nondecreasing; its difference quotients on either side of a point are nonnegative, and their common limit $f''(x)$ is therefore nonnegative.

Assume $f''\ge0$ on $I$. Applying [L2] to $f'$ shows that $f'$ is nondecreasing.

By [L1], the conclusion of step 1.2 makes $f$ convex, while step 1.1 proves the reverse implication.

For $n=0$, the only weights are $0$ and $1$, and the asserted inequalities are equalities.

Assume the assertion at $n$. At $n+1$, an even numerator $2j$ reduces to the induction hypothesis; an odd numerator $2j+1$ is the midpoint of the adjacent weights $j/2^n$ and $(j+1)/2^n$, so [L1] followed by the induction hypothesis proves the assertion.

The base and successor steps establish the assertion for every natural $n$ and every $0\le k\le2^n$.

For every $n$, [L3] applied to $2^n\lambda$ supplies $k_n$ with $k_n/2^n\le\lambda<(k_n+1)/2^n$; the elementary induction $2^n\ge n+1$ and [L2] show $k_n/2^n\to\lambda$.

Apply [L1] at the dyadic weight $k_n/2^n$ and let $n\to\infty$. Continuity of $f$ at $\lambda x+(1-\lambda)y$ and ordinary limit laws give the convexity inequality at $\lambda$.

At $\lambda=0$ and $\lambda=1$ the inequality is equality; with step 2.1 this proves convexity for every weight in $[0,1]$.

For $N=1$, the sole weight is $1$, so the two sides are both $f(x_1)$.

Assume the assertion for $N-1$ terms. If $\lambda_N=1$, all earlier nonnegative weights vanish and the assertion is immediate; otherwise put $s=1-\lambda_N>0$ and normalize the earlier weights as $\mu_i=\lambda_i/s$.

The induction hypothesis bounds $f(\sum_{i<N}\mu_i x_i)$ by $\sum_{i<N}\mu_i f(x_i)$; applying [L1] to this point and $x_N$ with weights $s,\lambda_N$ gives the asserted $N$-term inequality.

Fix $r>0$ as in [L2], and suppose for contradiction that some $y\in I$ has $f(y)<f(c)$.

Choose $0<\lambda<\min\{1,r/|y-c|\}$ when $y\ne c$ and put $z=(1-\lambda)c+\lambda y$; then $z\in I$ and $|z-c|<r$.

Convexity gives $f(z)\le(1-\lambda)f(c)+\lambda f(y)<f(c)$, contradicting the local-minimum inequality.

Suppose distinct points $x,y\in I$ are both global minimizers.

Their midpoint lies in $I$, and [L1] gives $f((x+y)/2)<(f(x)+f(y))/2=f(x)$.

This value is below a global minimum, a contradiction; hence there are at most one such point.

On the side where $f''\ge0$, [L1] makes $f$ convex; on the side where $f''\le0$, apply [L1] to $-f$ to make $f$ concave.

If the signs occur in the opposite order, the same argument interchanges the two sides.

In either case the assumed continuity at $c$ and the change of convexity/concavity meet [L2]'s definition.