# Is it a named result (or consequence thereof) that decreasing functions integrable against $e^{kx}$ decay faster than $e^{-kx}$?

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?

• 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 $x\rightarrow\infty$. Commented Mar 16, 2022 at 20:40
• @CarloBeenakker I specified that $f$ is decreasing. Commented Mar 16, 2022 at 20:51

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 $$$$f(x+)=\int_{(x,\infty)} \mu_f(dt)$$$$ for all $$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)$$, \begin{aligned} J(y)&:=\int_y^\infty f(x)e^{kx}\,dx \\ &=\int_y^\infty f(x+)e^{kx}\,dx \\ &=\int_y^\infty e^{kx}\,dx\, \int_{(x,\infty)} \mu_f(dt) \\ &=\int_{(y,\infty)}\, \mu_f(dt) \int_y^t e^{kx}\,dx \\ &\ge\int_{[y+1,\infty)}\, \mu_f(dt) \int_y^t e^{kx}\,dx \\ &\ge\int_{[y+1,\infty)}\, \mu_f(dt) \int_y^{y+1} e^{kx}\,dx \\ &=f((y+1)-) e^{k(y+1)}h(k) \\ &\ge f(y+1) e^{k(y+1)}h(k), \end{aligned} \tag{1}\label{1} where $$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.
• 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. Commented Mar 16, 2022 at 22:06