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Hello my question is really inspired by the following example, but also many others which are similar. Compute $\displaystyle\lim_{n \to \infty} \sqrt{n} \int_{-1}^{1} e^{\frac{-nx^{2}}{2}}f(x) \ \mathrm dx $

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.

When we have a limit of an integral to compute, and we change variables to get it to a more tractable form, and the change of variables is a function of n and x, say here its $t= \sqrt{n} x$, thereby eliminating the $nx^2$ from that term and sort of covering it by t, and then we take the limit of the whole function as $ n\to \infty$, aren't we assuming t is independent of n when it is dependent on n?

Thanks in advance.

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Upon enforcing the substitution $\displaystyle t=\sqrt n x$, for $\displaystyle x\in [-1,1]$, $\displaystyle t\in [-\sqrt n,\sqrt n]$ becomes a "dummy" index that spans $[-\sqrt n,\sqrt n]$.

It is, therefore, not dependent on $n$, but rather it spans a domain that depends on $\displaystyle n$.

Proceeding, we have

$$\begin{align} \sqrt{n}\int_{-1}^1 e^{-\frac12nx^2}f(x)\,dx&=\int_{-\sqrt n}^{\sqrt n} e^{-\frac12t^2}f(t/\sqrt n)\,dt\\\\ &=\int_{-\infty}^\infty e^{-\frac12t^2}f(t/\sqrt n))\xi_{[-\sqrt n,\sqrt n]}(t)\,dt \end{align}$$

Inasmuch as $f\in C[-1,1]$, it is bounded there. Then, we have

$$\left|e^{-\frac12t^2}f(t/\sqrt n))\xi_{[-\sqrt n,\sqrt n]}(t)\right|\le ||f||_{\infty}e^{-\frac12 t^2}\in L^1$$

where $||f||_{\infty}=\sup_{x\in [-1,1]}|f(x)|$.

Hence, we have

$$\begin{align} \lim_{n\to \infty}\sqrt{n}\int_{-1}^1 e^{-\frac12nx^2}f(x)\,dx&=\int_{-\infty}^\infty \lim_{n\to \infty}\left(e^{-\frac12t^2}f(t/\sqrt n))\xi_{[\sqrt n,\sqrt n]}(t)\right)\,dt\\\\ &=f(0)\int_{-\infty}^\infty e^{-\frac12t^2}\,dt\\\\ &=\sqrt{2\pi }f(0) \end{align}$$

And we are done!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Feb 19 '17 at 20:18
  • $\begingroup$ Thanks! Your answer is informative, I am wondering on a couple of things though, first of all, how exactly is the continuity used? Is that because otherwise supremum of f might be infinite? If we were instead told f is bounded would the result have been the same? $\endgroup$ – user172377 Feb 19 '17 at 23:34
  • $\begingroup$ You're welcome. My pleasure. Continuity permits us to write $\lim_{n\to\infty}f(t/\sqrt n)=f(0)$ along with assuring that $f$ attains a maximum on a closed bounded interval. $\endgroup$ – Mark Viola Feb 19 '17 at 23:43
  • $\begingroup$ ah yes! thank you, so if we had simply that f is bounded we would have to leave it as $\lim_{n\to\infty}f(t/\sqrt n)$ which may or may not exist in that case? And I will upvote soon! $\endgroup$ – user172377 Feb 19 '17 at 23:47
  • $\begingroup$ Yes, if $f$ had a jump discontinuity at $0$, for example, then the limt would not be $f(0)$. $\endgroup$ – Mark Viola Feb 20 '17 at 0:21

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