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

$$\tau_\epsilon(x)= \begin{cases}\frac{1}{\varepsilon}, & -\frac{\varepsilon}{2}<x<\frac{\varepsilon}{2} \\ 0, & \text { otherwise. }\end{cases}$$

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.

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

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

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

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;


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

Here's my attempt for 2.

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

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

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;

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

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

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

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.

  • $\begingroup$ Your proof for 1 is fine. See also math.stackexchange.com/questions/3071085 $\endgroup$ Feb 11 at 15:43
  • $\begingroup$ 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? $\endgroup$ Feb 11 at 15:48
  • $\begingroup$ 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. $\endgroup$
    – Adam
    Feb 11 at 16:17
  • $\begingroup$ 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? $\endgroup$
    – JDoe2
    Feb 11 at 16:39
  • $\begingroup$ 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. $\endgroup$
    – Adam
    Feb 11 at 16:42


You must log in to answer this question.