Compute $\displaystyle\lim_{n \to \infty} \sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ \mathrm dx $
where $f:[-1,1] \to \mathbb{R}$ is a continuous function.
I have no clue how to do it... Any hints?
Compute $\displaystyle\lim_{n \to \infty} \sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ \mathrm dx $
where $f:[-1,1] \to \mathbb{R}$ is a continuous function.
I have no clue how to do it... Any hints?
Following the hints, by letting $t=\sqrt{n}x$, we have that $$\sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ dx=\int\limits_{\mathbb{R}} e^{-\frac{t^{2}}{2}}f\left(\frac{t}{\sqrt{n}}\right)I_{[-\sqrt{n},\sqrt{n}]}(t)\ dt$$ Now for all $t\in\mathbb{R}$, $$F_n(t):=e^{-\frac{t^{2}}{2}}f\left(\frac{t}{\sqrt{n}}\right)I_{[-\sqrt{n},\sqrt{n}]}(t)\to F(t):=e^{-\frac{t^{2}}{2}}f(0)$$ and $$|F_n(t)|\leq Me^{-\frac{t^{2}}{2}}$$ where $M=\sup_{x\in[-1,1]}|f(x)|$.
Since $Me^{-\frac{t^{2}}{2}}$ is integrable over $\mathbb{R}$, by the Dominated Convergence Theorem $$\int_{\mathbb{R}}F_n(t) \ dt\to \int_{\mathbb{R}}F(t) \ dt=f(0)\int_{\mathbb{R}}e^{-\frac{t^{2}}{2}}\ dt=f(0)\cdot \sqrt{2\pi}.$$
Since $f$ is continuous on $[-1,1]$, the integrand is also continuous on $[-1,1]$. By mean-value theorem there is some $c \in [-1,1]$ such that the integral $= 2\exp (-nc^{2}/2)f(c)$, where $2$ is the length of the interval. Can you finish it from here?