Finding distributional limit

Question

How to find $\lim_{\varepsilon\rightarrow 0+}f_{\varepsilon}$ in $D'(R)$, if $f_\varepsilon$ is defined as: $f_\varepsilon(x)=\frac{1}{\varepsilon^3}$ for $x\in(0,\varepsilon)\cup(2\varepsilon,3\varepsilon)$, $f_\varepsilon(x)=\frac{-2}{\varepsilon^3}$ for $x\in(\varepsilon,2\varepsilon)$, $f_\varepsilon(x)=0$ for $x\notin(0,3\varepsilon)$. Thanks in advance.

Answer 1

To preface this, I'm not that experienced with rigorous distribution-theoretic proofs so I can't give all details but hopefully you can fill in the things I am missing. $f_{\varepsilon}$ is defined for all $\varepsilon$ so it's a well-defined function. The issue is then what the limit of $f_{\varepsilon}$ is. Let's write the following:

$$\langle f_{\varepsilon},\phi\rangle = \int_{-\infty}^{\infty} f_{\varepsilon}(x)\phi(x)dx = \int_{0}^{\varepsilon}\frac{1}{\varepsilon^3}\phi(x)dx + \int_{\varepsilon}^{2\varepsilon}\frac{-2}{\varepsilon^3}\phi(x)dx + \int_{2\varepsilon}^{3\varepsilon}\frac{1}{\varepsilon^3}\phi(x)dx$$

Let's make a change of variable $x\rightarrow x-\varepsilon$ and $x\rightarrow x-2\varepsilon$ in the second and third integrals, respectively:

$$\langle f_{\varepsilon},\phi\rangle = \frac{1}{\varepsilon}\int_{0}^{\varepsilon}\frac{\phi(x)-2\phi(x+\varepsilon)+\phi(x+2\varepsilon)}{\varepsilon^2}dx .$$

The integrand (in the limit that $\varepsilon\rightarrow 0$) becomes $\phi''(x)$ (you can see this from simple Taylor series arguments) and the outer limit looks like an application of the fundamental theorem of calculus. So I think we have that:

$$\lim_{\varepsilon\rightarrow 0}\langle f_{\varepsilon},\phi\rangle = \phi''(0).$$

I think this indicates that the distributional limit of $f_{\varepsilon}$ is $\delta''$. Someone else can perhaps clear up any mistakes I've made or fill in any missing details.