Proof on delta sequences

Question

In a coursetext of mine there is a theorem that says: when $f(x)$ is locally integrable ($f(x)\in\text{L}^1_{\text{loc}}(\mathbb{R}^d$), $f(x)\geq0 \,\forall x$ and $\int f(x) dx=1$ then it follows that $\lim_{n\rightarrow \infty} n^df(nx)=\delta(x)$.

This I get, but there is an exercise which asks us to prove that in the one-dimensional case the $f(x)\geq 0$ can be dropped.

I should define the function $$F_n(x)=\int_{-\infty}^x f_n(x) dx$$ ($f_n(x)=nf(nx)$). I am able to prove that $F_n(x)$ converges to the Heaviside function $H$ pointwise. Now I wish to use Lesbesgues Dominated Convergence theorem to prove that $F_n\rightarrow H$, but in order to do this I first have to proof that $\forall n\, \forall x \,|F_n(x)|<c$, where $c$ is some constant.

Now $$\left|\int_{-\infty}^x f_n(t) dt\right|=\left|\int_{-\infty}^{nx} f(t) dt\right|$$ So I have to prove that $\left|\int_{-\infty}^{s} f(t) dt\right|$ is smaller then some constant for all $s$, but I can't seem to find any reason why that should be. Could anyone offer me some help, be gentle, I'm only an engineer.

Answer 1

Define a function $G(x) = \int_0^x f(s)ds$. The existence of the integral $\int_{\Bbb R}f(x)dx $ means that the both limits $\lim_{x\to +\infty} G(x)$ and $\lim_{x\to -\infty} G(x)$ exist and are finite.

Now $G(x)$ is a continuous function which admits finite limits on $\pm\infty$, therefore it is bounded (easy exercise).

Finally, $$ \int_{-\infty}^{s} f(t) dt = \lim_{x\to -\infty} G(x) + G(s),$$ i.e. a finite constant plus a bounded function, therefore, you have the boundedness.