Consider the following: fix a function $\bar{b} : \mathbf{R}\_+ \to [0, \infty]$, and define


\begin{align}
\mathcal{S} \left( \bar{b} \right) := \left\{ b : \mathbf{R}\_+ \to [0, \infty] \, \text{s.t.} \, b \leq \bar{b} \, \text{pointwise} \right\}.
\end{align}


Recall the set of totally-monotone functions $\mathcal{B}$, defined as the set of functions $b$ so that for each nonnegative integer $k$, it holds that
\begin{align}
\text{for} \, t \in \mathbf{R}\_+, \quad (-1)^k \left( \frac{\mathrm{d}}{\mathrm{d}t} \right)^{(k)}
b \geq 0 \quad
\end{align}


Treating $\bar{b}$ as fixed, I would like to find a totally-monotone function $\tilde{b}$ which is a valid upper bound for all functions in $\mathcal{S} \left( \bar{b} \right) \cap \mathcal{B}$.


That is, given the set of totally-monotone functions which are upper-bounded by $\bar{b}$, I would like to be able to say that the same functions can also be upper-bounded by a totally-monotone function $\tilde{b}$.


Ideally, it would also be the case that this new bound is *at least* as tight as the original bound, i.e.


\begin{align}
0 \leq \tilde{b} \leq \bar{b}
\end{align}


Given that the set of totally monotone functions is a convex polytope, I am hopeful that there is a relatively simple argument which shows this (e.g. perhaps only using the convex / polytope structure), but I have not been able to crack it myself.



Can upper bounds on totally monotone functions be taken (WLOG) to be themselves totally monotone?
$\newcommand{\tb}{\tilde b}\newcommand{\bb}{\bar b}\newcommand{\S}{\mathcal S}\newcommand{\B}{\mathcal B}\newcommand{\T}{\mathcal T}$Note that any totally-monotone function is nonnegative and nonicreasing.
So, trivially, the constant function $\tb:=\bb(0)$ is a totally-monotone majorant of all functions $b\in\T(\bb):=\S(\bb)\cap\B$.




---


However, in general there is no totally-monotone majorant $\tb$ of all $b\in\T(\bb)$
such that $\tb\le\bb$.


Indeed, consider the following example. Let
\begin{equation}
 \bb:=1\_{[0,1)}+e^{-1}1\_{[1,\infty)}. 
\end{equation}
Suppose that there is a totally-monotone majorant $\tb$ of all $b\in\T(\bb)$
such that $\tb\le\bb$.


The constant function $e^{-1}$ is in $\T(\bb)$ and hence $\tb\ge e^{-1}$. One the other hand, $\bb=e^{-1}$ on $[1,\infty)$ and $\tb\le\bb$, whence $\tb\le e^{-1}$ on $[1,\infty)$. So, $\tb=e^{-1}$ on $[1,\infty)$.


By Bernstein's theorem,
\begin{equation}
 \tb(x)=\int\_{[0,\infty)}e^{-t x}\mu(dt)
\end{equation}
for some (finite nonnegative) measure $\mu$ and all real $x\ge0$. Using now analytic continuation and the condition $\tb=e^{-1}$ on $[1,\infty)$, we see that $\tb=e^{-1}$ on $[0,\infty)$.


But this contradicts the condition that $\tb$ majorizes the function $b\_1\in\T(\bb)$ given by the formula $b\_1(x):=e^{-x}$ for all real $x\ge0$. $\quad\Box$.




---


The OP asked in a comment if the additional assumption that all functions involved are positive and decreasing to $0$ at infinity can help.


The answer, however, is still negative. Indeed, let us modify the above example as follows. For real $x\ge0$, let
\begin{equation}
 \bb(x):=1(x<1)+e^{-(1+x)/2}1(x\ge1). 
\end{equation}
Suppose that there is a totally-monotone majorant $\tb$ of all $b\in\T(\bb)$
such that $\tb\le\bb$.


The function $[0,\infty)\ni x\mapsto e^{-(1+x)/2}$ is in $\T(\bb)$ and hence $\tb(x)\ge e^{-(1+x)/2}$ for all $x\in[0,\infty)$. One the other hand, $\bb(x)=e^{-(1+x)/2}$ for $x\in[1,\infty)$ and $\tb\le\bb$, whence $\tb(x)\le e^{-(1+x)/2}$ for $x\in[1,\infty)$. So, $\tb(x)=e^{-(1+x)/2}$ for $x\in[1,\infty)$. So, by Bernstein's theorem and analytic continuation, $\tb(x)=e^{-(1+x)/2}$ for $x\in[0,\infty)$.


But this contradicts the condition that $\tb$ majorizes the function $b\_1\in\T(\bb)$ given by the formula $b\_1(x):=e^{-x}$ for all real $x\ge0$. $\quad\Box$.


