Prove $ \lim_{x\to \infty} f^{'}_n(x)=f^{'}(x) $, being $ f_n(x)=\frac{\sqrt{(1-x^2)^{n^2+1}}}{n} , \forall{x}\in{[-1,1], \forall{n}\in{\mathbb{N}}}$

Question

I've got $ f_n(x)=\frac{\sqrt{(1-x^2)^{n^2+1}}}{n} , \forall{x}\in{[-1,1], \forall{n}\in{\mathbb{N}}}$
I've analyzed that $ f_n(x) $ is punctually and uniformly convergent in $ [-1,1] $.
Now, I have to prove that $ \lim_{x\to \infty} f^{'}_n(x)=f^{'(x)} $ being $ f(x) $ the punctual limit of $ f^{'}_n(x) $ in $ [-1,1] $

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If I'm not wrong, what I need to do to prove this, is to see if the following statements are true:
$ 1) $ $f_n(x)$ is differentiable in $[-1,1]$. This is true.
$ 2) $ $ \exists t\in [-1,1] $ where {$ f_n(t) $} $_{n\in\mathbb{N}} $ is convergent. This is true (I've proved it)
$ 3) $ {$ f^{'}_n(t) $} $_{n\in\mathbb{N}} $ is uniformly convergent in $[-1,1]$.

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So what is left is to prove the third point.

My first problem is that I don't know really good how to do the third point. I mean, I've calculated $ f^{'}_n(x)=\frac{-x(n^2+1)(1-x)^{n^2}}{n\sqrt{(1-x^2)^{n^2+1}}} $ but I don't know how to calculate that limit.

Anyway, I've supposed that the limit is $ 0 $ so that I can continue.

I also have to calculate $ d_n=\sup|f_n(x)| $ and for that I have to differentiate $ f^{'}_n(x) $ and calculate the maximum. I don't know if I've done this step well but I obtain that the maximum is $ x= \frac{-1}{\sqrt{n}} $. But when I replace it in the equation, the limit of $ d_n $ gives me $ \infty $, and that means that is not uniformly convergent in $[-1,1]$ but I know that it is. I don't know what I'm doing wrong...

Answer 1

We have $f_n(x) = n^{-1} (1-x^2)^{(n^2+1)/2}$. Termwise, $f_n'(x) = \frac{-(n^2+1)}{n} x (1-x^2)^{(n^2-1)/2}$. There are three cases:

• $x=0$. Clearly $f_n'(x)=0\to0$.
• $x=\pm 1$. Clearly $f_n'(\pm 1)=0\to 0$.
• $|x|<1$. Denote $c=1-x^2$. Then for $n>2$ we have $$ |f_n'(x)|\leq \frac{n^2+1}{n} c^{(n^2-1)/2}\leq 2n c^{n}\to 0 $$So the derivative is uniformly convergent to $0$ for any $x\in [-1,1]$.

For the second part, note that $(1-x^2)^{(n^2+1)/2}\leq 1$, with equality only at $x=0$.