Convexity

Put $\lambda =(z-y)\mathrm{/}(z-x)\in (0,1)$; 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 $M:=\max \{\mathrm{\mid }s(u,a)\mathrm{\mid },\mathrm{\mid }s(b,v)\mathrm{\mid }\}$, step 1.1 yields $\mathrm{\mid }f(y)-f(x)\mathrm{\mid }\le M\mathrm{\mid }y-x\mathrm{\mid }$ for $x<y$; symmetry gives the same estimate for all $x,y\in [a,b]$, which is the Lipschitz condition.

Given $c\in \mathrm{int}I$, choose such an interval $[a,b]$ containing $c$ in its interior; the estimate in step 2.1 gives the $\epsilon$--$\delta$ condition at $c$ by taking $\delta =\epsilon \mathrm{/}M$ 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 $f_{-}^{\mathrm{\prime }}(c)$ and $f_{+}^{\mathrm{\prime }}(c)$.

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

Put $g(c):=f_{+}^{\mathrm{\prime }}(c)$. The order chain in [L1] gives $g(u)\le g(v)$ whenever $u<v$, so $g$ is nondecreasing.

For $u<c$, [L1] gives $g(u)\le f_{-}^{\mathrm{\prime }}(c)$. Conversely, if $a<u<c$, [L1] gives $s(a,u)\le f_{-}^{\mathrm{\prime }}(u)\le g(u)$. Letting $u\to c^{-}$, continuity from [L3] gives ${\lim }_{u\to c^{-}}s(a,u)=s(a,c)$; then letting $a\to c^{-}$ gives ${\lim }_{u\to c^{-}}g(u)=f_{-}^{\mathrm{\prime }}(c)$. Since $g(c)=f_{+}^{\mathrm{\prime }}(c)$, 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 $(f(c)-f(x))\mathrm{/}(c-x)\le f_{-}^{\mathrm{\prime }}(c)\le m$.

If $x>c$, [L1] gives $m\le f_{+}^{\mathrm{\prime }}(c)\le (f(x)-f(c))\mathrm{/}(x-c)$.

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 $f^{\mathrm{\prime }}(u)\le f^{\mathrm{\prime }}(v)$, so $f^{\mathrm{\prime }}$ is nondecreasing.

Assume instead that $f^{\mathrm{\prime }}$ is nondecreasing. For a fixed $c$, the derivative of $g(x)=f(x)-f^{\mathrm{\prime }}(c)x$ 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 $f(x)\ge f(c)+f^{\mathrm{\prime }}(c)(x-c)$ 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], $f^{\mathrm{\prime }}$ is nondecreasing on $I$.

Thus $f^{\mathrm{\prime }}$ is monotone, so [L2] gives continuity of $f^{\mathrm{\prime }}$.

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

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

Assume $f^{\mathrm{\prime }\mathrm{\prime }}\ge 0$ on $I$. Applying [L2] to $f^{\mathrm{\prime }}$ shows that $f^{\mathrm{\prime }}$ 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 $j\mathrm{/}{2}^n$ and $(j+1)\mathrm{/}{2}^n$, so [L1] followed by the induction hypothesis proves the assertion.

The base and successor steps establish the assertion for every natural $n$ and every $0\le k\le 2^n$.

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

Apply [L1] at the dyadic weight $k_n\mathrm{/}{2}^n$ and let $n\to \mathrm{\infty }$. 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 ${\mu }_i={\lambda }_i\mathrm{/}s$.

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

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 $f((x+y)\mathrm{/}2)<(f(x)+f(y))\mathrm{/}2=f(x)$.

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

On the side where $f^{\mathrm{\prime }\mathrm{\prime }}\ge 0$, [L1] makes $f$ convex; on the side where $f^{\mathrm{\prime }\mathrm{\prime }}\le 0$, 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.