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
On
You may try to rewrite first and then take the limit $$ h\int_0^\infty e^{-hx}f(x)\,dx-f(0)=\int_0^\infty e^{-t}\left[f\left(\frac{t}{h}\right)-f(0)\right]\,dt. $$
On
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, $$ \lim_{h\rightarrow\infty}h\int_{0}^{\infty}e^{-hx}f(x)dx =f(0). $$
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 $\int \limits _0 ^\infty \Bbb e ^{-y} f(0) \Bbb d y = f(0)$.