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Apologies if this question is too basic for MO.

I think it should be the case that

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 $x \to \infty$.

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. $\ \square$

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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  • $\begingroup$ what if $f$ is not continuous? say, $f(x)=1/x$ if $x\in\mathbb{N}$ and $f(x)=0$ otherwise; the integral vanishes but $f(x)e^{kx}$ does not go to $0$ as $x\rightarrow\infty$. $\endgroup$ Commented Mar 16, 2022 at 20:40
  • $\begingroup$ @CarloBeenakker I specified that $f$ is decreasing. $\endgroup$ Commented Mar 16, 2022 at 20:51

1 Answer 1

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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.


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.

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  • $\begingroup$ Beautiful. I think your argument can simplify quite a lot: With $h(x):=e^{kx}$, we have $J(y)=\int_y^\infty fh \geq \int_y^{y+1} fh \geq f(y+1) \int_y^{y+1} h$. Then the rest of the argument proceeds as you've written. $\endgroup$ Commented Mar 16, 2022 at 22:06
  • $\begingroup$ @JulianNewman : Thank you for your comment. Indeed, this is much simpler. $\endgroup$ Commented Mar 17, 2022 at 3:45

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