Distributional Limit $T_{g_{n}} \rightarrow \delta_{0}$

Question

Consider $$ \psi(x)=\left\{\begin{array}{ll} e^{-\frac{1}{1-x^{2}}}, & |x|<1 \\ 0, & |x| \geqslant 1 \end{array}\right. $$ and $$ g_{n}(x)=\frac{\psi(n x)}{\displaystyle\int_{-\infty}^{\infty} \psi(n x) \: \mathrm{d} x} $$ for $n \in \mathbb{N}$. I need to prove that $T_{g_{n}} \rightarrow \delta_{0}$ in $\mathscr{D}^{\prime}$ for $n \rightarrow+\infty$. ($T_{g_{n}}$ denotes the regular distribution induced by $g_n$).

I have tried to write down the action of $T_{g_n}$ on a test function $\phi(x)$ as: $$\langle T_{g_n}, \phi(x)\rangle=\int_{-\frac 1n}^{\frac 1n} \frac{e^{-\frac{1}{1-n^2x^{2}}}}{\displaystyle\int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2y^{2}}} \: \mathrm{d} y} \phi(x) \: \mathrm{d}x = \frac{1}{\displaystyle\int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2y^{2}}} \: \mathrm{d} y} \int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2x^{2}}} \phi(x) \: \mathrm{d} x $$ But I really don't know what to do from here. Any ideas?

Answer 1

HINT:

Enforce the substitution $x\mapsto x/n$ to find that

$$\int_{-1/n}^{1/n}e^{-1/(1-n^2x^2)}\,\phi(x)\,dx=\frac1n \int_{-1}^1 e^{-1/(1-x^2)}\,\phi(x/n)\,dx$$

Now, divide by $\frac1n \int_{-1}^1 e^{-1/(1-x^2)}\,dx$ and apply the Dominated Convergence Theorem.

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NOTE:

As mentioned by @paulgarrett in a comment, there is nothing particularly special about the Bump function $e^{-1/(1-x^2)}$ on $[-1,1]$. For any non-negative function $\psi(x)\in C_C$ with $\psi(0)>0$, we see that for any $\phi\in C_C^\infty$

$$\lim_{n\to \infty}\frac{\int_{-\infty}^\infty \psi(nx)\phi(x)\,dx}{\int_{-\infty}^\infty \psi(nx)\,dx}=\phi(0)$$

Answer 2

Given $\varepsilon>0$, there is $\delta>0$ such that $x|<\delta$ implies $|\phi(x)-\phi(0)|<\varepsilon$

$$\left|\int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2x^{2}}} (\phi(x)-\phi(0))\,\mathrm{d} x\right|\leq\varepsilon\int_{-\frac 1n}^{\frac 1n} e^{-\frac{1}{1-n^2x^{2}}} \,dx $$

Thus

$$ \Big|\frac{1}{\int\psi(nx)\,dx} \int\psi(nx)(\phi(x)-\phi(0))\,dx\Big|\leq \varepsilon $$ for all $n$ large enough.

Notice that it is enough to consider $\phi\in\mathcal{C}_b(\mathbb{R})$. This means that $\mu_n(dx)=\frac{1}{\int\psi(xn)\,dx}\psi(nx)\,dx$ converges weakly (as probability measures) to $\delta_0$.