Prove that : $\lim_{n\to +\infty}\int_{-\infty}^{+\infty}f_{n}(x)\phi(x)\,dx=\phi(0)$

Question

Let defined for $n≥1$ :

$$f_{n}(x)=n\left(1-\frac{x^{2}}{n}\right)^{n^{3}}$$

Then prove that :

$$\displaystyle\lim_{n\to +\infty}\int_{-\infty}^{+\infty}f_{n}(x)\phi(x)dx=\phi(0)$$

Then prove that :

$$\lim_{n\to +\infty}\langle f_{n},\phi\rangle=\sqrt{\pi}\langle\delta,\phi\rangle$$

where $\phi\in D$ mean $\operatorname{supp}\phi\subset [-a,a]$

I have tried substation $y=\frac{x}{\sqrt{k}}$

I know that $\int_{\Bbb R}\delta\phi dx=\phi(0)$

But I don't know how I started in first limits ?

Because the last question from first question ?

Answer 1

Take $x=\frac{t}{n}$ thus $dx=\frac{dt}{n}$ so the integral becomes $$\int_{\Bbb{R}}(1-\frac{t^2}{n^3})^{n^3}\phi(\frac{t}{n})dt$$

Note that $g_n(t) \to e^{-t^2}\phi(0)$

Also since $\phi$ is a Schwarz function,then it is bounded by some $M>0$ and from the inequality $1-x \leq e^{-x},\forall x \geq 0$ we have that $$|g_n(t)|\leq |\phi(x)|e^{-\frac{t^2}{n^3}n^3}\leq Me^{-t^2} \in L^1(\Bbb{R})$$ so by Dominated Convergence you have that the sequence of integrals converges to $$\phi(0)\int_{\Bbb{R}}e^{-t^2}dt=\sqrt{\pi}\phi(0)$$