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?

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.