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Uniform convergence of: $$f_n(x):={(1-\frac{x^2}{n})^n}$$

(a) In a closed interval $[-L, L]$ as $L>0$.

(b) In $\mathbb{R}$.

My try:

There is a pointwise convergence to $e^{-x^2}$.

Obviously, there is no uniform convergence in $\mathbb{R}$ as $\lim\limits _{x\to \infty }|f_n(x)|=\infty$.$\;$ Thus, $\sup \limits _{x\in\mathbb{R}}|f_n(x)-e^{-x^2}|= \infty$

Can anyone give a direction for a closed interval?

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Using the basic inequalities $e^y \geqslant 1 +y$ and $e^{-y} \geqslant 1 - y$ for $0 \leqslant y \leqslant 1$ we can show that for $x^2 < n$,

$$\left(1- \frac{x^2}{n}\right)^n \leqslant e^{-x^2}\leqslant \left(1+ \frac{x^2}{n}\right)^{-n}.$$

Applying the Bernoulli inequality $(1 - x^4/n^2)^n \geqslant 1 - x^4/n,$ we get

$$0 \leqslant e^{-x^2} - \left(1- \frac{x^2}{n}\right)^n = e^{-x^2}\left[1 - e^{x^2}\left(1- \frac{x^2}{n}\right)^{n}\right]\\ \leqslant e^{-x^2}\left[1 - \left(1+ \frac{x^2}{n}\right)^{n}\left(1- \frac{x^2}{n}\right)^{n}\right]\\= e^{-x^2}\left[1 - \left(1- \frac{x^4}{n^2}\right)^{n}\right]\leqslant e^{-x^2}\frac{x^4}{n}.$$

For $x \in [-L,L]$, the RHS is bounded by $L^4/n$, which proves uniform convergence.

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  • $\begingroup$ You know that $e^y > 1 +y$. Take $y = x^2/n$ and we get $e^{x^2/n} > 1+x^2/n$ which implies the inequality on the right. $\endgroup$ – RRL Apr 30 '18 at 17:44
  • $\begingroup$ You also have $e^{-y} > 1 - y$ which leads to the left inequality. Do I need to add this to the answer? $\endgroup$ – RRL Apr 30 '18 at 17:46
  • $\begingroup$ I assume you are familiar with $e^y > 1 + y$ which follows from the Taylor expansion for y > 0$. $\endgroup$ – RRL Apr 30 '18 at 17:47
  • $\begingroup$ Thank you very much. No need to add it. $\endgroup$ – Um Shmum Apr 30 '18 at 17:49
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    $\begingroup$ @Um Shmum: You're welcome. $\endgroup$ – RRL Apr 30 '18 at 17:54
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To find $\sup \limits _{x\in\mathbb{[-L,L]}}|f_n(x)-e^{-x^2}|$; supremum will be exist for some $ {x\in\mathbb{[-L,L]}}$.Suppose $x=a$ then $\lim\limits_{n\rightarrow\infty} \sup |f_n(x)-e^{-x^2}|=0$.Hence uniformly convergent

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  • $\begingroup$ This just shows pointwise convergence, for $x=a$, unless you can somehow show that it's the same $a$ for all $n\in \mathbb{N}$ $\endgroup$ – Um Shmum Apr 30 '18 at 17:52
  • $\begingroup$ I take that a for which supremum willbe exist and after calculating supremum I take limit $\endgroup$ – Epsilon Delta Apr 30 '18 at 19:15

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