limit with Riemann integral

Question

Assume $f\colon\mathbb{R}\to\mathbb{R}$ is Riemann integrable on $[0,A]$ for every $A>0$ and $\lim_{x\to \infty}f(x)=1$. Prove that $$ \lim_{t\to 0^+}\ t\!\int_{0}^{\infty} e^{-tx}f(x) dx\ =\ 1 $$

Answer 1

Since $\int_0^\infty te^{-tx}\, dx = 1$ for $t > 0$, we can write

$$\int_0^\infty te^{-tx}f(x)\, dx - 1 = \int_0^\infty te^{-tx}(f(x) - 1)\, dx$$

for $t > 0$. So it suffices to show $\limsup_{t\to 0^+}\, \lvert\int_0^\infty te^{-tx}(f(x) - 1)\, dx\rvert = 0$. Let $\epsilon > 0$. Since $\lim_{x\to \infty} f(x) = 1$, there exists a positive number $A$ such that for all $x \ge A$, $|f(x) - 1| < \epsilon$. Hence

\begin{align}\left\lvert \int_0^\infty te^{-tx}(f(x) - 1)\, dx\right\rvert &\le \int_0^\infty te^{-tx}|f(x) - 1|\, dx\\ & = \int_0^A te^{-tx}|f(x) - 1|\, dx + \int_A^\infty te^{-tx}|f(x) - 1|\, dx\\ &\le t\int_0^A |f(x) - 1|\, dx + \epsilon\int_A^\infty te^{-tx}\, dx\\ &= t\int_0^A |f(x) - 1|\, dx + \epsilon\cdot e^{-tA} \end{align}

Since the last expression tends to $\epsilon$ as $t \to 0^+$, we obtain

$$\limsup_{t\to 0^+} \left\lvert \int_0^\infty te^{-tx}(f(x) - 1)\, dx\right\rvert \le \epsilon$$

Since $\epsilon$ was arbitrary, the result follows.

Answer 2

There was a correct proof given, using Dominated Convergence. The answerer deleted it, I don't know why. In any case, that previously if briefly appearing argument can easily be converted to one using no measure theory.

Note that $f$ is bounded, since it has a limit at infinity and is bounded on $[0,A]$ for every $A$.

First make that change of variables: $$t\int e^{-tx}f(x)\,dx=\int e^{-x}f(x/t)\,dt.$$Now let $\epsilon>0$. Since $f$ is bounded there exist $\delta>0$ and $A<\infty$ so that $$\int_0^\delta e^{-x}|f(x/t)|\,dx +\int_A^\infty e^{-x}|f(x/t)|\,dx<\epsilon$$for every $t$. But $f(x/t)\to1$ uniformly on $[\delta,A]$ as $t\to0$, hence $$\lim_{t\to0}\int_\delta^Ae^{-x}f(x/t)\,dx=\int_\delta^Ae^{-x}\,dx.$$

(That will count as a complete proof for many readers; others should take it as a hint and fill in the details themselves... note that "$\limsup$" thingie in one of the other answers. )

Answer 3

For any $\epsilon\gt0$, there is an $x_\epsilon$ so that for $x\ge x_\epsilon$, we have $\left|\,f(x)-1\,\right|\le\epsilon$. $$ \begin{align} \left|\,\lim_{t\to0^+}\,t\!\int_0^\infty\!e^{-tx}f(x)\,\mathrm{d}x-1\,\right| &=\left|\,\lim_{t\to0^+}\,t\!\int_0^\infty\!e^{-tx}(f(x)-1)\,\mathrm{d}x\,\right| \tag{1}\\ &\le\lim_{t\to0^+}\,t\!\int_0^{x_\epsilon}\!e^{-tx}\left|f(x)-1\right|\,\mathrm{d}x\tag{2}\\ &+\lim_{t\to0^+}\,t\!\int_{x_\epsilon}^\infty\!e^{-tx}\left|f(x)-1\right|\,\mathrm{d}x\tag{3}\\[3pt] &\le x_\epsilon\sup_{[0,x_\epsilon]}\left|f(x)-1\right|\lim_{t\to0^+}t\tag{4}\\ &+\lim_{t\to0^+}\epsilon t\int_0^\infty e^{-tx}\,\mathrm{d}x\tag{5}\\[6pt] &=\epsilon\tag{6} \end{align} $$ Explanation:
$(1)$: $t\int_0^\infty e^{-tx}\,\mathrm{d}x=1$
$(2)$: triangle inequality
$(3)$: triangle inequality
$(4)$: from $(2)$, $e^{-tx}\left|f(x)-1\right|\le\sup\limits_{[0,x_\epsilon]}\left|f(x)-1\right|$ on $[0,x_\epsilon]$
$(5)$: from $(3)$, $\left|f(x)-1\right|\le\epsilon$ for $x\ge x_\epsilon$
$(6)$: evaluate the simple limits in $(4)$ and $(5)$

Since $(6)$ is true for any $\epsilon\gt0$, we have $$ \lim_{t\to0^+}\,t\!\int_0^\infty\!e^{-tx}f(x)\,\mathrm{d}x=1\tag{5} $$