Apologies if this question is too basic for MO.


I think it should be the case that



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> for any decreasing $f \colon [A,\infty) \to [0,\infty)$ and $k \geq 0$, if $\int\_A^\infty f(x) e^{kx} \, dx < \infty$ then $f(x)e^{kx} \to 0$ as $x \to \infty$.
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**Proof** [I think]**.** Write $g(x)=f(x)e^{kx}$. If $f$ is differentiable, argue by contrapositive that if $g$ is integrable but $\{x:g(x) \geq \varepsilon\}$ is unbounded for some $\varepsilon>0$, then $f$ is not decreasing: Take sequences $x\_n,y\_n \to \infty\,$ with $\,x\_n - y\_n \downarrow 0$ such that $g(x\_n) = \varepsilon$, $g \leq \varepsilon$ on $[y\_n,x\_n]$ and $g(y\_n) \leq \frac{\varepsilon}{2}$; then we can find $z\_n \in (y\_n,x\_n)$ such that $g'(z\_n) \geq \frac{\varepsilon}{2(x\_n-y\_n)}$ and hence $f'(z\_n)=e^{-kz\_n}(g'(z\_n)-kg(z\_n)) \geq \varepsilon e^{-kz\_n}\!\left(\frac{1}{2(x\_n-y\_n)}-k\right)$, which is clearly positive for sufficiently large $n$, and so $f$ is not decreasing.


If $f$ is non-differentiable, let $\varphi$ be a nonnegative unit-integral $C^1$ bump function supported on an interval $[0,\delta]$, and define $\tilde{f} \colon [A+\delta,\infty) \to [0,\infty)$ by $\tilde{f}= f \ast \varphi$; then writing $\tilde{g}(x)=\tilde{f}(x)e^{kx}$, we have
$$ g(x) \leq \tilde{g}(x) \leq e^{k\delta}g(x-\delta)\text{,} $$
and so the differentiable case applied to $\tilde{f}$ gives the result. $\ \square$



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> If the result is correct, it seems like it should be a standard result: either a result with a name, or an immediate special case of a result with a name. Is it so?
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Is it a named result (or consequence thereof) that decreasing functions integrable against $e^{kx}$ decay faster than $e^{-kx}$?
Here is a simple proof. Since $f$ is nonnegative and decreasing, there is a limit $c:=\lim\_{x\to\infty}f(x)\ge0$, and $f\ge c$. So, $\infty>\int\_A^\infty f(x)e^{kx}\,dx
\ge\int\_A^\infty ce^{kx}\,dx=\infty$ if $c>0$. So, $c=0$ and hence
\begin{equation}
 f(x+)=\int\_{(x,\infty)} \mu\_f(dt)
\end{equation}
for all $x\in[A,\infty)$, where $\mu\_f$ is the Lebesgue--Stieltjes measure based on the decreasing function $f$, so that $\mu\_f((a,b])=f(a+)-f(b+)$ for all $a$ and $b$ in $[A,\infty)$ such that $a\le b$.


Hence, by the Tonelli theorem, for any $y\in[A,\infty)$,
\begin{equation}
\begin{aligned}
 J(y)&:=\int\_y^\infty f(x)e^{kx}\,dx \\ 
 &=\int\_y^\infty f(x+)e^{kx}\,dx \\ 
 &=\int\_y^\infty e^{kx}\,dx\, \int\_{(x,\infty)} \mu\_f(dt) \\ 
 &=\int\_{(y,\infty)}\, \mu\_f(dt) \int\_y^t e^{kx}\,dx \\ 
 &\ge\int\_{[y+1,\infty)}\, \mu\_f(dt) \int\_y^t e^{kx}\,dx \\ 
 &\ge\int\_{[y+1,\infty)}\, \mu\_f(dt) \int\_y^{y+1} e^{kx}\,dx \\ 
 &=f((y+1)-) e^{k(y+1)}h(k) \\ 
 &\ge f(y+1) e^{k(y+1)}h(k), 
\end{aligned}
\tag{1}\label{1}
\end{equation}
where $h(k):=(1-e^{-k})/k$ if $k>0$, with $h(0):=1$, so that $h(k)>0$ for each real $k\ge0$.


The condition $\int\_A^\infty f(x)e^{kx}\,dx<\infty$ implies $J(y)\to0$ and hence, by \eqref{1}, $f(y+1) e^{k(y+1)}\to0$ as $y\to\infty$. Thus, the desired conclusion holds.




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Since this proof is very simple and transparent, it is unlikely that it is published anywhere as a theorem. It is quite possible that it appears in some, perhaps quite unlikely, papers as a lemma. Such statements are much easier to prove than to find in literature.


