Proof of Supporting Lines, Composition and Jensen's Inequality for a Convex Lipschitz Integrand

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here. The field axioms and the rules for adding inequalities, for multiplying or dividing them by positive real numbers, for multiplying them by nonnegative real numbers, and for handling absolute values, from Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field and Properties of the Absolute Value in an Ordered Field, are used without further mention.

Step 1 (Claim 1). Let $t_{0}\in[0,\infty)$.

Case $t_{0}=0$. Take $s=0$. For $t\in[0,\infty)$ the Lipschitz bound of Convex Lipschitz Integrands §integrand, applied with $a=0\le b=t$, together with $\Phi(0)=0$, gives $0\le\Phi(t)$, which is the inequality $\Phi(t)\ge\Phi(0)+0\cdot(t-0)$.

Case $0<t_{0}$. First we prove the three-slope inequality: for $t,u\in[0,\infty)$ with $t<t_{0}<u$,

Put $\lambda=(u-t_{0})(u-t)^{-1}$. Since $0<u-t_{0}<u-t$, we have $0\le\lambda\le1$, and $1-\lambda=(t_{0}-t)(u-t)^{-1}$, so that $\lambda t+(1-\lambda)u=\bigl((u-t_{0})t+(t_{0}-t)u\bigr)(u-t)^{-1}=t_{0}(u-t)(u-t)^{-1}=t_{0}$. The convexity inequality of Convex Lipschitz Integrands §integrand, applied with $a=t$, $b=u$ and the parameter $\lambda$, gives $\Phi(t_{0})\le\lambda\Phi(t)+(1-\lambda)\Phi(u)$. Multiplying by $u-t>0$ and writing $u-t=(u-t_{0})+(t_{0}-t)$ on the left,

and dividing by the positive number $(u-t_{0})(t_{0}-t)$ gives (1).

Let $A$ be the set of the real numbers $(\Phi(t_{0})-\Phi(t))(t_{0}-t)^{-1}$ with $t\in[0,\infty)$ and $t<t_{0}$. It is nonempty, since $t=0$ qualifies, and by (1) with $u=t_{0}+1$ the number $\Phi(t_{0}+1)-\Phi(t_{0})$ is an upper bound of $A$. Hence $A$ has a least upper bound $s$ by The Real Numbers: Standing Notation and Background §bounds, in the sense of Upper Bound and Least Upper Bound. Let $t\in[0,\infty)$. If $t<t_{0}$, then $(\Phi(t_{0})-\Phi(t))(t_{0}-t)^{-1}\le s$, since $s$ is an upper bound of $A$; multiplying by $t_{0}-t>0$ gives $\Phi(t_{0})-\Phi(t)\le s(t_{0}-t)$, that is, $\Phi(t)\ge\Phi(t_{0})+s(t-t_{0})$. If $t=t_{0}$, the inequality is an equality. If $t_{0}<t$, then by (1) with $u=t$ the number $(\Phi(t)-\Phi(t_{0}))(t-t_{0})^{-1}$ is an upper bound of $A$, so $s\le(\Phi(t)-\Phi(t_{0}))(t-t_{0})^{-1}$, $s$ being the least upper bound; multiplying by $t-t_{0}>0$ gives $\Phi(t)\ge\Phi(t_{0})+s(t-t_{0})$. This proves claim 1.

Step 2 (Measurability and pointwise bounds in claim 2). Let $h$ be as in claim 2. Define $\tilde\Phi:\mathbb{R}\to\mathbb{R}$ by $\tilde\Phi(r)=\Phi(\max\{r,0\})$. For real $r,r'$ let $a$ and $b$ be the smaller and the larger of the nonnegative numbers $\max\{r,0\}$ and $\max\{r',0\}$; the Lipschitz bound of Convex Lipschitz Integrands §integrand gives

the last step by the elementary inequality $|\max\{r,0\}-\max\{r',0\}|\le|r-r'|$, checked by distinguishing the signs of $r$ and $r'$. Thus $\tilde\Phi$ is Lipschitz with constant $L$ as a map from the real line $(\mathbb{R},d_{\mathbb{R}})$ with its absolute-value metric to itself, hence continuous by A Lipschitz Map is Uniformly Continuous. By claim 3 of Borel Measurability and Bounded Integration on a Metric Space, $\tilde\Phi$ is measurable with respect to the Borel $\sigma$-algebra of $(\mathbb{R},d_{\mathbb{R}})$, which is the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ by claim 2 of that lemma. The map $h$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(\mathbb{R})$ by Measure Spaces and the Lebesgue Integral: Standing Notation §measurable, and $\Phi\circ h=\tilde\Phi\circ h$ because $h\ge0$; hence $\Phi\circ h$ is measurable by claim 4 of Borel Measurability and Bounded Integration on a Metric Space. For $x\in X$, the Lipschitz bound of Convex Lipschitz Integrands §integrand with $a=0\le b=h(x)$ and $\Phi(0)=0$ gives $0\le\Phi(h(x))\le L\,h(x)$.

Step 3 (Integrals in claim 2). Suppose in addition that $h$ is integrable. Since $h\ge0$, its positive part is $h$ and its negative part is $0$, so by Integrable Function and the Lebesgue Integral its real integral $\int_{X}h\,d\mu$ equals the integral of $h$ as a nonnegative measurable function in the sense of Measure Spaces and the Lebesgue Integral: Standing Notation §integral; in particular $0\le\int_{X}h\,d\mu<\infty$. The function $\Phi\circ h$ is nonnegative and measurable by Step 2, and by claim 1 of Linearity and Monotonicity of the Lebesgue Integral (the multiple $Lh$ and monotonicity),

As $|\Phi\circ h|=\Phi\circ h$, the criterion recorded in Measure Spaces and the Lebesgue Integral: Standing Notation §integral shows that $\Phi\circ h$ is integrable, and, as for $h$, its real integral is the integral just bounded, which lies in $[0,\infty)$. This proves claim 2.

Step 4 (Claim 3). Suppose $\mu(X)=1$ and let $h$ be as in claim 2 and integrable. By Step 3, $t_{0}=\int_{X}h\,d\mu$ belongs to $[0,\infty)$. By claim 1 choose a real $s$ with $\Phi(t)\ge\Phi(t_{0})+s(t-t_{0})$ for every $t\in[0,\infty)$. The indicator $\mathbf{1}_{X}$ is a nonnegative simple function with $\int_{X}\mathbf{1}_{X}\,d\mu=\mu(X)=1$ by The Integral of an Indicator Function is the Measure of the Set; hence, as in Step 3, it is integrable with real integral $1$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, the function $g=(\Phi(t_{0})-s\,t_{0})\mathbf{1}_{X}+s\,h$ is integrable and

For every $x\in X$, taking $t=h(x)$ in the supporting-line inequality gives $\Phi(h(x))\ge\Phi(t_{0})+s\,(h(x)-t_{0})=g(x)$. Since $\Phi\circ h$ is integrable by claim 2, the monotonicity statement in claim 2 of Linearity and Monotonicity of the Lebesgue Integral yields

which is claim 3.