Prove that a function is a nascent delta function

Question

I need to prove that $$f_{n}(x)=\begin{cases}n^2(x-1)+n & x \in \left(1-\dfrac{1}{n},1\right]\\[6pt] n^2(1-x)+n&x \in \left(1,1+\dfrac{1}{n}\right) \\[6pt] 0 & x \in \left(0,1-\dfrac{1}{n}\right]\cup\left[1+\dfrac{1}{n},2\right) \end{cases} $$ is Dirac Delta distribution at $1$ (i.e. is a nascent delta function). This is: \begin{equation} \lim_{n\to\infty}\int_0^2f_n(x)\varphi(x)\,dx=\delta_1=\varphi(1) \end{equation}

Answer 1

Let's extend $f_n$ to also be $0$ outside $(0,\,2)$. We want to prove $g_n(x):=f_n(x+1)$ satisfies $\lim_{n\to\infty}\int_{\Bbb R} g_n(x) \varphi(x) \, dx=\varphi(0)$. But $g_n(x)=n(1-|nx|)=ng_1(nx)$, so it suffices to verify $g_1$ is a PDF.

Answer 2

The test function $\varphi$ is continuous. So for $\varepsilon>0$ we can find $N$ so big that for all $n\ge N,$ for all $x\in(1-\tfrac 1 n, 1 + \tfrac 1 n), \quad$ $\varphi(1)-\varepsilon<\varphi(x)< \varphi(1)+\varepsilon.$ In that case, $$ \int\limits_{\mathbb R} f_n(x) (\varphi(1)-\varepsilon)\, dx \le \int\limits_{\mathbb R} f_n(x)\varphi(x)\, dx \le \int\limits_{\mathbb R} f_n(x)(\varphi(1)+\varepsilon)\, dx $$ The values of the first and third integrals above are $\varphi(1)\pm\varepsilon.$