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reframing question in terms of uniform bounds

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edited May 25, 2022 at 15:38

I am interested in a function $b : \mathbf{R}_+ \to [0, \infty]$ withConsider the following properties: fix a function $\bar{b} : \mathbf{R}_+ \to [0, \infty]$, and define

I know a priori that $b$ is a `totally monotone' function, i.e. for each nonnegative integer $k$, it holds that \begin{align} (-1)^k \left( \frac{\mathrm{d}}{\mathrm{d}t} \right)^{(k)} b \geq 0 \quad \text{on}\, \mathbf{R}_+\end{align}

I am told that some function $\bar{b}$ is a global upper bound on $b$. A priori, I know relatively little about $\bar{b}$; let's say that I know that it is nonnegative and decreases to $0$.

\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}

I would like to argueRecall the set of totally-monotone functions $\mathcal{B}$, defined as the set of functions $b$ so that for each nonnegative integer $k$, given this informationit 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 can formwould like to find a new upper bound on $b$, saytotally-monotone function $\tilde{b}$ (againwhich is a function), such thatvalid upper bound for all functions in $\mathcal{S} \left( \bar{b} \right) \cap \mathcal{B}$.

$\tilde{b}$ is always at least as good of an upper bound, i.e. $\tilde{b} \leq \bar{b}$, and,

$\tilde{b}$ is also a totally monotone function.

EssentiallyThat is, given the set of totally-monotone functions which are upper-bounded by $\bar{b}$, I wantwould like to be able to say "WLOG, if I have anthat the same functions can also be upper bound on-bounded by a totally monotone-monotone function $\tilde{b}$.

Ideally, then I can takeit would also be the case that upperthis new bound to be totally monotone"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.

I am interested in a function $b : \mathbf{R}_+ \to [0, \infty]$ with the following properties:

I know a priori that $b$ is a `totally monotone' function, i.e. for each nonnegative integer $k$, it holds that \begin{align} (-1)^k \left( \frac{\mathrm{d}}{\mathrm{d}t} \right)^{(k)} b \geq 0 \quad \text{on}\, \mathbf{R}_+\end{align}

I am told that some function $\bar{b}$ is a global upper bound on $b$. A priori, I know relatively little about $\bar{b}$; let's say that I know that it is nonnegative and decreases to $0$.

I would like to argue that, given this information, I can form a new upper bound on $b$, say $\tilde{b}$ (again a function), such that

$\tilde{b}$ is always at least as good of an upper bound, i.e. $\tilde{b} \leq \bar{b}$, and,

$\tilde{b}$ is also a totally monotone function.

Essentially, I want to be able to say "WLOG, if I have an upper bound on a totally monotone function, then I can take that upper bound to be totally monotone".

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}

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asked May 25, 2022 at 13:54

πr8

Can upper bounds on totally monotone functions be taken (WLOG) to be themselves totally monotone?

I am interested in a function $b : \mathbf{R}_+ \to [0, \infty]$ with the following properties:

I know a priori that $b$ is a `totally monotone' function, i.e. for each nonnegative integer $k$, it holds that \begin{align} (-1)^k \left( \frac{\mathrm{d}}{\mathrm{d}t} \right)^{(k)} b \geq 0 \quad \text{on}\, \mathbf{R}_+\end{align}
I am told that some function $\bar{b}$ is a global upper bound on $b$. A priori, I know relatively little about $\bar{b}$; let's say that I know that it is nonnegative and decreases to $0$.

I would like to argue that, given this information, I can form a new upper bound on $b$, say $\tilde{b}$ (again a function), such that

$\tilde{b}$ is always at least as good of an upper bound, i.e. $\tilde{b} \leq \bar{b}$, and,
$\tilde{b}$ is also a totally monotone function.

Essentially, I want to be able to say "WLOG, if I have an upper bound on a totally monotone function, then I can take that upper bound to be totally monotone".