# Prove properties rigourously without using the delta function

Suppose, I have a $$C^1$$ functions $$f,g:\mathbb{R}\rightarrow\mathbb{R}$$. Define the unit bump function $$\tau_\epsilon(x)$$ as:

$$\tau_\epsilon(x)= \begin{cases}\frac{1}{\varepsilon}, & -\frac{\varepsilon}{2}

Without referencing the Dirac delta function, how can I prove the following two properties:

$$1. \quad \lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} f(x) \tau_\epsilon(x)dx=f(0)$$ $$2. \quad \lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} f(x) \tau_\epsilon(g(x)) \mathrm{d} x=\sum_i \frac{f\left(x_i\right)}{\left|g^{\prime}\left(x_i\right)\right|} \quad \text{ where } x_i \text{ are the roots of }g$$

The reason I don't want to use the Dirac delta function is that I have seen a lot of conflicting discussion (like discussed here and here) and am trying to convince myself that the above statements hold by returning to basics. A lot of the discussion around the Dirac delta seems to be centred on the space of $$C^\infty_C$$ functions, but I do not want to assume anything about infinite differentiability or compact supports.

Here is my attempt for 1.

$$\lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} f(x) \tau_\epsilon(x)dx$$

$$=\lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} f(x) \frac{1}{\epsilon}\mathbb{1}\bigg[-\frac{\epsilon}{2}

$$=\lim_{\epsilon \rightarrow 0}\int_{-\epsilon/2}^{\epsilon/2} \frac{1}{\epsilon} f(x) dx$$

As $$f$$ is continuous its integral $$F$$ exists so;

$$=\lim_{\epsilon \rightarrow 0}\frac{1}{\epsilon} \bigg(F\bigg(\frac{\epsilon}{2}\bigg)-F\bigg(-\frac{\epsilon}{2}\bigg)\bigg) dx$$

By the fundamental theorem of calculus;

$$=F'(0)=f(0).$$

Does this hold? Or have I missed some assumptions needed in this argument?

Here's my attempt for 2.

$$\lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} f(x) \tau_\epsilon(g(x)) \mathrm{d} x$$

$$=\lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} f(x) \frac{1}{\epsilon}\mathbb{1}\bigg[-\frac{\epsilon}{2}

By a Taylor approximation (which we can do because $$g$$ is $$C^1$$) around a root of $$g$$, $$x_i$$, we get that;

$$g(x)=g(x_i) + g'(x_i)(x_i-x) + O(x_i-x) = g'(x_i)(x_i-x) + O(x_i-x)$$

So we now have;

$$=\lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} f(x) \frac{1}{\epsilon}\mathbb{1}\bigg[-\frac{\epsilon}{2}

$$=\lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} \frac{f(x)}{|g'(x_i)|} \frac{|g'(x_i)|}{\epsilon}\mathbb{1}\bigg[-\frac{\epsilon}{2|g'(x_i)|}<(x_i-x) + O(x_i-x)<\frac{\epsilon}{2|g'(x_i)|}\bigg]dx$$

$$=\lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty} \frac{f(x)}{|g'(x_i)|} \tau_{\frac{\epsilon}{|g'(x_i)|}}\bigg((x_i-x) + O(x_i-x)\bigg)$$

This looks kind of right... but I think I've gone wrong somewhere as I only have one root and I don't know what to do with this remainder term. I feel like I could have missed some assumptions here and I'm not sure how valid my approach of substituting the Taylor expansion was here.

• Your 2 and its proof are dubious. What if, for instance, $g$ has only one root $x_0$ but $g'(x_0)=0?$ Or if it has a converging sequence of roots? Feb 11 at 15:48
• A rough proof for 2: assume $g$ has finitely many roots $x_i$, $g'(x_i) \ne 0$, and $g$ is bounded away from $0$ at $\pm \infty$. Then for $\epsilon$ small, the integral is over disjoint intervals $A_i$ containing each $x_i$. Arguing as in 1, we can replace $f(x)$ with $f(x_i)g'(x)/g'(x_i)$ on $A_i$; the result then follows from integration by substitution.
• Thanks @Adam - I'm a bit confused by the substitution though, is the partition based on $\epsilon$? Why was the bound away from zero needed too? Feb 11 at 16:39
• The partition is fixed, it's just some disjoint small-enough intervals that cover the integral for small-enough $\epsilon$. The bound away from zero ensures we don't have to worry what happens as $x \to \pm \infty$; it's an alternative to assuming $f$ has compact support.