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Let a sequence of functions $(f_n)_{n\in \mathbb{N}^*}$ such that $f_n(x)=\left(\cos\dfrac{x}{\sqrt{n}}\right)^n$ defined on $\mathbb{R}$

  1. Show the pointwise convergence on $\mathbb{R}$

  2. Show that the sequence converges uniformly on $[-A,A]$ with $A>0$


Pointwise convergence

I fund that $f_n(x)\longrightarrow e^{-x^2/2}$

Uniform convergence

I must show that $||f_n(x)-e^{-x^2/2}||_{\infty}\underset{n\to +\infty}{\longrightarrow }0$

My first idea (*to be honest I've got no overview) is to write :

$\bigg|f_n(x)-e^{-x^2/2}\bigg|$, and I tried to find a bound.

As $1-\dfrac{u^2}{2}\le\cos u\le 1-\dfrac{u^2}{2}+\dfrac{u^4}{24}\iff 1-\dfrac{x^2}{2n}\le \cos\dfrac{x}{\sqrt{n}}\le 1-\dfrac{x^2}{2n}+\dfrac{x^4}{24n^2}$

Let $g_1(x)=1-\dfrac{x^2}{2n}$ and $g_2(x)=1-\dfrac{x^2}{2n}+\dfrac{x^4}{24n^2}$ and $M_1, M_2$ such that :

For $M_1$ that is but for $M_2$ I can find a finite bound $$|g_1(x)|<M_1\quad \text{and}\quad |g_2(x)|<M_2$$

I repeat again I haven't got any overview about how to attack this problem

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    $\begingroup$ It's not a duplicate, because the uniform convergence hasn't been proven $\endgroup$ – Stu Dec 1 '18 at 21:09
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    $\begingroup$ I reopened after closing as a duplicate of this. The answer there does prove non-uniform convergence on $\mathbb{R}$, but fails to address uniform convergence on a compact interval. There probably is another duplicate somewhere. Note that this can be proved by squeezing with your estimates and using the inequality $0 \leqslant e^{-y} - \left(1 - \dfrac{y}{n}\right)^n \leqslant \dfrac{e^{-y}y^2}{n}$, $\endgroup$ – RRL Dec 2 '18 at 0:06
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$$\cos(x/\sqrt{n})^n=(1-x^2/(2n)+a(x) x^4/(24n^2))^n$$

where $-1 \leq a(x) \leq 1$. This is the Lagrange remainder form of Taylor expansion.

Using the assumption $x \in [-A,A]$ gives that $\cos(x/\sqrt{n})^n$ is always between $(1-x^2/(2n)+A^4/(24n^2))^n$ and $(1-x^2/(2n)-A^4/(24n^2))^n$. Now you are set up to use a squeezing argument and just need to be careful to select a $N(\varepsilon)$ that doesn't depend on $x$ (but will depend on $A$).

One way to do this squeezing argument is to note that $$(1-x^2/(2n)+A^4/(24n^2))^n \\ =\exp(n \log(1-x^2/(2n)+A^4/(24n^2)) \\ <\exp(-x^2/2)$$

using the Taylor expansion of $\ln(1+x)$. Then you do something similar but slightly more complicated for the lower bound.

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