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How can I prove the uniform convergence for these two tasks:

  1. $${\frac{1}{(1+ {\frac{y^2}{n}})^n}} ⇉ e^{-y^2} $$ I understand that the function on the right is the limit function, because: $$\lim _{{x\to \infty }}\left(1+{\frac {1}{x}}\right)^{x}=e.$$ but how can I prove exactly uniform convergence?
  2. $$∫_0^∞ \frac{dy}{(1+ {\frac{y^2}{n}})^n} ⇉$$
Just_A_User
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Arsen Tatraev
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  • $(1+y^2/n)^n\geq 1+y^2$ since $(a+b)^n\geq a^n+na^{n-1}b$ for $a,b>0$ (take $a=1$ and $b=y^2/n$ Thus $f_n(y)=\frac{1}{(1+y^2/n)^n}\leq \frac{1}{1+y^2}\in L_1$. Dominated convergence gives part (2), that is $\int f_n\rightarrow\int e^{-y^2}$ – Mittens Jun 28 '20 at 10:41
  • For (1) this may help: https://math.stackexchange.com/questions/2566469/uniform-bounds-for-1-y-nn-exp-y – Mittens Jun 28 '20 at 11:01
  • @OliverDiaz The DCT is likely beyond the scope of the OP's toolkit, especially in light of the tag "uniform convergence" and the first part of the question regarding the UC of the stated limit. I've posted a solution that addresses the issue of UC. Let me know your thoughts. ;-) – Mark Viola Jun 30 '20 at 17:05
  • @Mark Viola: I like the proof of part (1), although adding an argument that for fixed $a>0$, $x\mapsto (1+ (a/x))^x$ is monotone increasing may also help the OP. As for the second, I probably had bad memories in college when iterated limits (as in Apostol's section 9.39) were covered in class that when I learned dominated convergence I sweared never to go back. Be it as it may, your proof of (2) is also nice. Maybe just a quotation about the basic result on iterated limits would also illuminate things for the OP. but who know, may be the OP has a good understanding of that. – Mittens Jun 30 '20 at 17:22
  • @OliverDiaz I agree on the power of the DCT. Thank you for the suggestions. I've added a reference to my statement regarding the monotonicity of $\left(1+\frac{x}{n}\right)^n$ and a reference to the Moore-Osgood Theorem on iterated limits. Stay sate and healthy my friend. – Mark Viola Jun 30 '20 at 17:26

1 Answers1

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Let $f_n(x)$ be the sequence

$$f_n(y)=\left(1+\frac{y^2}{n}\right)^{-n}-e^{-y^2}\tag1$$

Since $\left(1+\frac{y^2}{n}\right)^{-n}$ monotonically decreases to $e^{-y^2}$ (See this answer as a reference), we see from $(1)$ that $f_n(y)\ge0$ for all $y\ne 0$.

Setting $f_n'(y)=0$, we find the maximum value of $f_n(y)$ occurs at some value $y^*>0$ (and $-y^*$) for which

$$e^{-{y^*}^2}=\left(1+\frac{{y^*}^2}{n}\right)^{-n-1}\tag2$$

Using $(2)$, reveals that the maximum value of $f_n(y)$ is given by

$$\begin{align} \max_y(f_n(y))&=f_n(\pm y^*)\\\\ &=\left(1+\frac{{y^*}^2}{n}\right)^{-n}-e^{-{y^*}^2}\\\\ &=e^{-{y^*}^2}\left(1+\frac{{y^*}^2}{n}-1\right)\\\\ &=\frac{{y^*}^2e^{-{y^*}^2}}n\\\\ &\le \frac1{ne} \end{align}$$

Hence, we find that

$$\sup_{y}\left|\left(1+\frac{y^2}{n}\right)^{-n}-e^{-y^2}\right|\le \frac{1}{ne}\to 0$$

and conclude that the sequence $\left(1+\frac{y^2}n\right)^n$ converges uniformly to $e^{-y^2}$ for $y\in \mathbb{R}$ as was to be shown!

Note that $f_n(y)$ is continuous in $y$ for all $n$ and that $\int_0^\infty f_n(y)\,dy$ converges uniformly for $n\ge 1$. Then, the Moore-Osgood Theorem guarantees that we can interchange limits so that

$$\lim_{n\to\infty}\lim_{L\to\infty}\int_0^L f_n(y)\,dy=\lim_{L\to\infty}\lim_{n\to\infty}\int_0^L f_n(y)\,dy$$

Inasmuch as $f_n(y)$ converges uniformly for all $y$, we have

$$\begin{align} \lim_{L\to\infty}\lim_{n\to\infty}\int_0^L f_n(y)\,dy&=\lim_{L\to\infty}\int_0^L \lim_{n\to\infty} f_n(y)\,dy\\\\ &=\lim_{L\to\infty}\int_0^L e^{-y^2}\,dy\\\\ &=\frac{\sqrt\pi}{2} \end{align}$$

Mark Viola
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