Proof of Convex Functions on an Open Interval: Monotone Slopes, Supporting Lines, Continuity, and the Supremum of Rational Supporting Lines

Throughout, for $x\in I$ and $y\in I$ with $y\neq x$ write

$$Q_{x}(y)=\frac{f(y)-f(x)}{y-x},\qquad\text{so that } Q_{x}(y)=Q_{y}(x);$$

and $x\mapsto c_{x}$, once $c_{x}$ exists, is a map $I\to\mathbb{R}$ given by an infimum (Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices (iv)), so the set in (v) is formed as written. Every $x\in I$ lies strictly between two points of $I$: since $I$ is open, there is a real $r>0$ with $B_{d_{\mathbb{R}}}(x,r)\subseteq I$, and $x-r/2$ and $x+r/2$ lie in this ball by Open Ball in a Metric Space (ii), as their distance to $x$ is $r/2<r$.

Step 1: increasing slopes. Let $s,t,u\in I$ with $s<t<u$, and put $\lambda=(u-t)/(u-s)$. Then $0<\lambda<1$, $1-\lambda=(t-s)/(u-s)$ and $\lambda s+(1-\lambda)u=t$, so convexity (Convex Functions on an Interval of ℝ (i), with $s$, $u$ and the weight $\lambda$) gives $f(t)\le\lambda f(s)+(1-\lambda)f(u)$. Subtracting $f(s)$, respectively $f(u)$, from both sides and rearranging,

$$f(t)-f(s)\le\frac{t-s}{u-s}\bigl(f(u)-f(s)\bigr),\qquad f(u)-f(t)\ge\frac{u-t}{u-s}\bigl(f(u)-f(s)\bigr).$$

Dividing the first inequality by $t-s>0$ and the second by $u-t>0$ gives the two inequalities of (i).

Step 2: the quotients at a point increase. Let $x\in I$ and $y,z\in I\setminus\{x\}$ with $y<z$. We show $Q_{x}(y)\le Q_{x}(z)$. If $x<y<z$, this is the left inequality of (i) for $(s,t,u)=(x,y,z)$. If $y<z<x$, it is the right inequality for $(y,z,x)$, since $Q_{x}(y)=(f(x)-f(y))/(x-y)$ and $Q_{x}(z)=(f(x)-f(z))/(x-z)$. If $y<x<z$, it is the outer inequality for $(y,x,z)$:

$$Q_{x}(y)=\frac{f(x)-f(y)}{x-y}\le\frac{f(z)-f(x)}{z-x}=Q_{x}(z).$$

Step 3: the infimum $c_{x}$. Let $x\in I$, and choose $s,t\in I$ with $s<x<t$, as above; so the set $R_{x}=\{Q_{x}(y):y\in I,\ y>x\}$ contains $Q_{x}(t)$, and by Step 2 every $s'\in I$ with $s'<x$ gives the lower bound $Q_{x}(s')$ of $R_{x}$. Hence $R_{x}$ has an infimum $c_{x}\in\mathbb{R}$, by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (viii); as a lower bound of $R_{x}$ it satisfies $c_{x}\le Q_{x}(t')$ for every $t'\in I$ with $t'>x$, and as the greatest lower bound it satisfies $Q_{x}(s')\le c_{x}$ for every $s'\in I$ with $s'<x$. Since $Q_{x}(s')=(f(x)-f(s'))/(x-s')$, this is (ii).

Step 4: supporting lines. Let $x,y\in I$. For $y=x$ there is nothing to prove. If $y>x$, then $c_{x}\le Q_{x}(y)$ by (ii), and multiplying by $y-x>0$ gives $f(y)-f(x)\ge c_{x}(y-x)$. If $y<x$, then $Q_{x}(y)\le c_{x}$ by (ii), and multiplying by $y-x<0$ gives $f(y)-f(x)\ge c_{x}(y-x)$ again. This is (iii).

Step 5: continuity. We use the $\varepsilon$-$\delta$ form of continuity on $I$ of The Real Line: Standing Notation and Background for Calculus on Intervals (vi). Let $x\in I$ and let $\varepsilon>0$ be real. Choose $s,t\in I$ with $s<x<t$, as above, and put $M=|Q_{x}(s)|+|Q_{x}(t)|$. For $y\in I$ with $s<y<t$ and $y\neq x$, Step 2 gives $Q_{x}(s)\le Q_{x}(y)\le Q_{x}(t)$, hence $|Q_{x}(y)|\le M$ and

$$|f(y)-f(x)|=|Q_{x}(y)|\,|y-x|\le M\,|y-x|,$$

which also holds for $y=x$. Put $\delta=\min\{x-s,\ t-x,\ \varepsilon/(M+1)\}>0$. Every $y\in I$ with $|y-x|<\delta$ satisfies $s<y<t$, so

$$|f(y)-f(x)|\le(M+1)\,|y-x|<(M+1)\,\delta\le\varepsilon .$$

Hence $f$ is continuous on $I$, which is (iv).

Step 6: the rational supporting lines. Let $y\in I$ and let $S_{y}=\{f(q)+c_{q}(y-q):q\in I\cap\mathbb{Q}\}$. By (iii), applied at $q$, every element of $S_{y}$ is at most $f(y)$. Choose $s\in I$ with $s<y$, as above; by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (x) there is $q\in\mathbb{Q}$ with $s<q<y$, and $q\in I$ by Intervals of the Real Line (i). So $S_{y}$ is nonempty and bounded above by $f(y)$; it has a supremum by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vii), and $\sup S_{y}\le f(y)$.

For the reverse inequality put $r=(s+y)/2$, so $s<r<y$ and $r\in I$, and

$$L=\frac{f(r)-f(s)}{r-s}.$$

For $q\in I$ with $r<q<y$, (ii) (with $s<q$) and then (i) (with $s<r<q$) give

$$c_{q}\ge\frac{f(q)-f(s)}{q-s}\ge L,\qquad\text{hence}\qquad f(q)+c_{q}(y-q)\ge f(q)+L(y-q)\ge f(q)-|L|\,(y-q).$$

Now let $\varepsilon>0$ be real. By (iv) there is a real $\delta>0$ such that $|f(q)-f(y)|<\varepsilon/2$ for every $q\in I$ with $|q-y|<\delta$. Put $a=\max\{r,\ y-\delta,\ y-\varepsilon/(2|L|+2)\}$, so $a<y$; by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (x) there is $q\in\mathbb{Q}$ with $a<q<y$, and $q\in I$ since $r<q<y$. Then $(|L|+1)(y-q)<\varepsilon/2$, and

$$f(q)+c_{q}(y-q)\ge f(q)-|L|\,(y-q)>\Bigl(f(y)-\frac{\varepsilon}{2}\Bigr)-\frac{\varepsilon}{2}=f(y)-\varepsilon .$$

So $\sup S_{y}>f(y)-\varepsilon$ for every real $\varepsilon>0$, hence $\sup S_{y}\ge f(y)$, and with the first part $f(y)=\sup S_{y}$. This is (v).