Proof of A Real Function with Nonnegative Second Derivative is Convex on an Interval

Order arithmetic is taken from Elementary Order Arithmetic in an Ordered Field (claim 1 strict compatibility with addition, claim 2 mixed transitivity, claim 4 sign reversal, claim 5 product of positive elements) and from Elementary Arithmetic in an Ordered Field (claim 3 translation, claim 5 multiplication by a nonnegative element), and the field axioms of the field $\mathbb{R}$ are used for rearrangement. Translating $\theta\le1$ by $-\theta$ gives $0\le1-\theta$.

Restriction. If $u,v\in J$ with $u<v$, then the closed interval $[u,v]$ is contained in $J$ because $J$ is order-convex, and the restrictions of $g$ and $g_1$ to $[u,v]$ are differentiable at every point $c$ of the open interval $(u,v)$, with the same derivative values: the condition of Derivative at an Interior Point for the function on $J$ quantifies over increments $h$ with $c+h\in J$, hence holds a fortiori for those with $c+h\in[u,v]$, and the values agree by Uniqueness of the Derivative at an Interior Point. Moreover the restriction of $g$ to $[u,v]$ is continuous on $[u,v]$: $g$ is differentiable at every point of $J$, hence continuous there by claim 1 of Sum and Product Rules for One-Dimensional Derivatives and Continuity, hence continuous on $J$ as a map into $(\mathbb{R},d_{\mathbb{R}})$ by Continuity of a Real Function Agrees with Metric Continuity on the Real Line, and restriction to a subset preserves this by claim 1 of Restriction of a Continuous Map, and Continuous Images of Compact Subsets. The same applies to $g_1$. Consequently Rolle's Theorem and the Mean Value Theorem on a Closed Interval may be applied to $g$ and to $g_1$ on any such $[u,v]$.

Step 1: $g_1$ is nondecreasing. Let $u,v\in J$ with $u<v$. By claim 2 of Rolle's Theorem and the Mean Value Theorem on a Closed Interval applied to $g_1$ there is $c\in(u,v)$ with $g_1'(c)\,(v-u)=g_1(v)-g_1(u)$. Translating $u<v$ by $-u$ gives $0<v-u$, and $0\le g_1'(c)$ by hypothesis, so multiplication by a nonnegative element gives $0\le g_1(v)-g_1(u)$, that is, $g_1(u)\le g_1(v)$ after translation by $g_1(u)$.

Step 2: membership. Let $s,t\in J$ and $\theta$ as in the statement, and put $w=(1-\theta)\,s+\theta\,t$. If $t\le s$, exchange the roles of $s$ and $t$ and replace $\theta$ by $1-\theta$, which leaves both $w$ and the asserted inequality unchanged; so assume $s\le t$. Then

$$w-s=\theta\,(t-s),\qquad t-w=(1-\theta)\,(t-s),$$

and $0\le t-s$, so both differences are nonnegative and $s\le w\le t$. Since $J$ is order-convex and $s,t\in J$, this gives $w\in J$.

Step 3: the inequality. If $s=t$ then $w=s$ and both sides equal $g(s)$. If $\theta=0$ then $w=s$ and the right-hand side is $g(s)$; if $\theta=1$ then $w=t$ and the right-hand side is $g(t)$. So assume $s<t$, $0<\theta$ and $\theta<1$; then $0<1-\theta$, and by step 2 together with $0<t-s$ we get $s<w<t$.

By claim 2 of Rolle's Theorem and the Mean Value Theorem on a Closed Interval applied to $g$ on $[s,w]$ and on $[w,t]$ there are $\xi_1\in(s,w)$ and $\xi_2\in(w,t)$ with

$$g(w)-g(s)=g'(\xi_1)\,(w-s)=g_1(\xi_1)\,\theta\,(t-s),\qquad g(t)-g(w)=g_1(\xi_2)\,(1-\theta)\,(t-s).$$

Since $\xi_1<w<\xi_2$, step 1 gives $g_1(\xi_1)\le g_1(\xi_2)$, hence $g_1(\xi_1)-g_1(\xi_2)\le0$ by translation. Multiplying by the nonnegative element $\theta\,(1-\theta)\,(t-s)$ and using sign reversal,

$$\theta\,(1-\theta)\,(t-s)\,\bigl(g_1(\xi_1)-g_1(\xi_2)\bigr)\le0 .$$

On the other hand, substituting the two displayed identities and regrouping,

$$\theta\,\bigl(g(w)-g(t)\bigr)+(1-\theta)\,\bigl(g(w)-g(s)\bigr)=\theta\,(1-\theta)\,(t-s)\,\bigl(g_1(\xi_1)-g_1(\xi_2)\bigr),$$

while the left-hand side equals $g(w)-\bigl((1-\theta)\,g(s)+\theta\,g(t)\bigr)$, because $\theta\,g(w)+(1-\theta)\,g(w)=g(w)$. Combining the last two displays gives

$$g(w)-\bigl((1-\theta)\,g(s)+\theta\,g(t)\bigr)\le0 ,$$

and translating by $(1-\theta)\,g(s)+\theta\,g(t)$ yields the asserted inequality.