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Let $f : [0, \infty) \to \Bbb R$ be bounded and continuous. Prove that $\lim \limits _{h \to \infty} h \int \limits _0 ^\infty e ^{-hx} f(x) \, d x = f(0)$.

Our intuition was to use l'Hospital's rule to try to find the limit, but it doesn't seem to work.

Thanks

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    $\begingroup$ For all finite $h$, the inside of the limit is $f(0)$. $\endgroup$
    – T.J. Gaffney
    Jul 29, 2015 at 16:55
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    $\begingroup$ @HarishChandraRajpoot: Please refrain from editing posts only to make minor modifications. Thank you. $\endgroup$
    – Alex M.
    Jul 29, 2015 at 17:43
  • $\begingroup$ @DanielFischer: I prefix the differential $\Bbb d$ and Euler's $\Bbb e$ by a "\Bbb" in order to visually distinguish them from functions or variables (i.e. common mathematical objects) $d$ or $e$. If I were a dictator, I would enforce this in every LaTeX piece of mathematical text. $\endgroup$
    – Alex M.
    Jul 29, 2015 at 18:02
  • $\begingroup$ See also: math.stackexchange.com/questions/1366970/… $\endgroup$
    – Martin Sleziak
    Jun 5, 2017 at 8:25

3 Answers 3

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Split the integral in two parts: $\int \limits _0 ^1$ and $\int \limits _1 ^\infty$.

For the second one, using the fact that $|f| \le M$, we obtain: $h \big| \int \limits _1 ^\infty \Bbb e ^{-hx} f(x) \Bbb d x \big| \le h \int \limits _1 ^\infty \Bbb e ^{-hx} |f(x)| \Bbb d x \le h \int \limits _1 ^\infty \Bbb e ^{-hx} M \Bbb d x = h M \frac {\Bbb e ^{-hx}} {-h} \big| _1 ^\infty = M \Bbb e ^{-h}$, which tends to $0$.

For the first one, we shall make the change of variable $y = hx$, thus obtaining: $h \int \limits _0 ^1 \Bbb e ^{-hx} f(x) \Bbb d x = \int \limits _0 ^h \Bbb e ^{-y} f(\frac y h) \Bbb d y$ which tends to $\int \limits _0 ^\infty \Bbb e ^{-y} f(0) \Bbb d y = f(0)$.

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    $\begingroup$ All true, but I suggest mentioning the rationale for the last step, which tacitly interchanged the order of the limit and the integral. Fairly straightforward but worth a mention. $\endgroup$
    – Mark Viola
    Jul 29, 2015 at 17:57
  • $\begingroup$ A +1 vote by the way ... $\endgroup$
    – Mark Viola
    Jul 29, 2015 at 19:59
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Because $h\int_{0}^{\infty}e^{-hx}dx=1$, you can write $$ \frac{1}{h}\int_{0}^{\infty}e^{-hx}f(x)dx-f(0)=\frac{1}{h}\int_{0}^{\infty}e^{-hx}\{f(x)-f(0)\}dx $$ If $M$ is a bound for $f$ on $[0,\infty)$ and $r > 0$, then, as $h\rightarrow\infty$, $$ \left|h\int_{r}^{\infty}e^{-hx}\{f(x)-f(0)\}dx\right| \le 2Mh\int_{r}^{\infty}e^{-hx}dx=2Mhe^{-h\delta}\rightarrow 0 $$ Let $\epsilon > 0$ be given. Because $f$ is continuous at $0$, there exists $\delta > 0$ such that $|f(x)-f(0)| < \epsilon/2$ whenever $0 \le x <\delta$. And, there exists $h_{0}$ such that $$ \left|h\int_{\delta/2}^{\infty}e^{-hx}\{f(x)-f(0)\}dx\right| < \epsilon/2 \mbox{ whenever } h > h_{0}. $$ Putting the pieces together: If $h > h_{0}$, then $$ \left|\frac{1}{h}\int_{0}^{\infty}e^{-hx}f(x)dx-f(0)\right| < \epsilon. $$ By the definition of limit, $$ \lim_{h\rightarrow\infty}h\int_{0}^{\infty}e^{-hx}f(x)dx =f(0). $$

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  • $\begingroup$ I believe that the endpoints of $\int \limits _{\frac \delta 2} ^\infty$ are not what you meant them to be, because that part of the proof should concern neighbourhoods of $0$, not of $\infty$. $\endgroup$
    – Alex M.
    Jul 30, 2015 at 8:38
  • $\begingroup$ What I have shown is as intended. The other piece $|h\int_{0}^{\delta/2}e^{-hx}(f(x)-f(0))dx| \le h\int_{0}^{\delta/2}e^{-hx}\frac{\epsilon}{2}dx < \frac{\epsilon}{2}$ for all $h > 0$. Choosing $\delta/2$ instead of $\delta$ is done because $|f(x)-f(0)| < \epsilon/2$ for $0 \le x < \delta$, and this avoids distracting discussion of the behavior of $|f(x)-f(0)|$ at $x=\delta$. Then you choose $h_{0}$ so that $h >h_{0}$ bounds the remaining piece by $\epsilon/2$, which can be done because of the exhibited limit property. This is quite rigorous. $\endgroup$
    – Disintegrating By Parts
    Jul 30, 2015 at 8:46
  • $\begingroup$ I didn't say it is not rigorous. I said that the original answer was incomplete, missing precisely the part that you have added in the above comment. $\endgroup$
    – Alex M.
    Jul 30, 2015 at 9:40
  • $\begingroup$ @AlexM. : I just wanted to let you know that I wasn't just waving my hands on that part. I figured I would let others think a little about it, because it's set up for the kill with everything else I had given, including how $\delta$ was defined, and it goes to the heart of the matter. $\endgroup$
    – Disintegrating By Parts
    Jul 30, 2015 at 9:46
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You may try to rewrite first and then take the limit $$ h\int_0^\infty e^{-hx}f(x)\,dx-f(0)=\int_0^\infty e^{-t}\left[f\left(\frac{t}{h}\right)-f(0)\right]\,dt. $$

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