Let $f \in L^1$, then there exists a function $\Theta$ having more than linear growth at infinity such that $\int \Theta(f(x))\,d\mu< \infty$

Question

Let $\mu_t$ be the positive measure on $\mathbb{R}^d$ for all $t\in \mathbb{R}^+$ and assume that $$\int_0^T\int_{\mathbb{R}^d}\frac{|b(x,t)|}{1+|x|}\,d\mu_t(x)\,dt\lt\infty.$$ Then there exists a convex nondecreasing function $\Theta: \mathbb{R}^+ \to \mathbb{\mathbb{R}}$ having more than linear growth at infinity such that $$\int_0^T\int_{\mathbb{R}^d}\frac{\Theta(|b(x,t)|)}{1+|x|}\,d\mu_t(x)\,dt \lt \infty.$$ The existence of $\Theta$ is ensured by the Dunford-Pettis Theorem.

The statement above is in the book "Transport equations and multi-D Hyperbolic conservation laws", page $13$. It seems that for $\mu$ a positive measure on $\mathbb{R}^d$ and $f \in L^1(d\mu)$, then there exists a convex nondecreasing function $\Theta: \mathbb{R}^+ \to \mathbb{\mathbb{R}}$ having more than linear growth at infinity such that $$\int_{\mathbb{R}^d}\Theta(f(x))\,d\mu\lt \infty.$$ The Dunford-Pettis Theorem states that every bounded set in $L^1(d\mu)$ is weakly compact in $L^1(d\mu)$ and I can't see why it implies the existence of $\Theta$. Any help and suggestions will by very appreciated.

Answer 1

I am not familiar with functional analysis, so let me tackle the question in a more elementary manner (or so I believe).

Let $\Theta : \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0} $ be convex and non-decreasing. We also assume $\Theta(0) = 0$ without losing the generality. Then the next result holds:

Theorem. There exists a Borel measure $\nu$ on $[0, \infty)$ such that $$\Theta(x) = \int_{[0, \infty)} (x - s)_+ \, \nu(\mathrm{d}s), $$ where $x_+ := \max\{0, x\}$ denotes the positive part of $x \in \mathbb{R}$.

Intuitively, $\nu(\mathrm{d}s) = \Theta''(s) \, \mathrm{d}s$ holds when $\Theta$ is of $C^2$. So, $\nu$ can be thought of as the distributional second derivative of $\Theta$. (More concretely, $\nu$ can be realized as the Lebesgue–Stieltjes measure associated to the right-derivative of $\Theta$.) We also have:

Corollary. $\Theta$ has super-linear growth if and only if $\nu$ is an infinite measure.

Proof. This is in immediate consequence of the computation $$ \lim_{x\to\infty} \frac{\Theta(x)}{x} = \lim_{x\to\infty} \int_{[0, \infty)} \left(1 - \frac{s}{x}\right)_+ \, \nu(\mathrm{d}s) = \int_{[0, \infty)} 1 \, \nu(\mathrm{d}s). $$

Now, consider a $\sigma$-finite measure $\mu$ on a measurable space $X$, and let $f \in L^1(\mu)$ be non-negative. Also, define the function $g : [0, \infty) \to [0, \infty)$ by

$$ g(s) = \int_{X} (f(x) - s)_+ \, \mathrm{d}x. $$

Since $f$ is integrable, $g$ is finite, non-increasing, and $g(s) \to 0$ as $s \to \infty$. Moreover, by the Fubini–Tonelli theorem,

\begin{align*} \int_{X} \Theta(f(x)) \, \mu(\mathrm{d}x) &= \int_{X} \int_{[0, \infty)} (f(x) - s)_+ \, \nu(\mathrm{d}s) \, \mu(\mathrm{d}x) \\ &= \int_{[0, \infty)} \int_{X} (f(x) - s)_+ \, \mu(\mathrm{d}x) \, \nu(\mathrm{d}s) \\ &= \int_{[0, \infty)} g(s) \, \nu(\mathrm{d}s). \end{align*}

With this formulation at our hand, we can now construct $\Theta$ that satisfies all the required properties. Indeed,

Construction. Choose $0 < s_1 < s_2 < \ldots$ such that $s_k \to \infty$ and $\sum_{k\geq 1} g(s_k) < \infty$. Then set $\nu$ and $\Theta$ as

$$ \nu = \sum_{k \geq 1} \delta_{s_k} \qquad\text{and}\qquad \Theta(x) = \int_{[0, \infty)} (x - s)_+ \, \nu(\mathrm{d}s) = \sum_{k \geq 1} (x - s_k)_+. $$

This $\nu$ is clearly an infinite measure, and so, $\Theta(x)$ has super-linear growth. (In fact, this can be directly proved from the representation $\Theta(x) = \sum_{k \geq 1} (x - s_k)_+$.) On the other hand,

$$ \int_{X} \Theta(f(x)) \, \mu(\mathrm{d}x) = \int_{[0, \infty)} g(s) \, \nu(\mathrm{d}s) = \sum_{k \geq 1} g(s_k) < \infty. $$