I am preparing for my exam and need help with the following tasks:
Let ${f_n}:[-1,1]\to \mathbb{R}$ be defined as $f_n=\frac{x}{1+n^2x^2}$
- Is ${f_n}$ pointwise/uniformly convergent? Specify the "limit-function". Determine the derivative of this function.
So I know that $fn$ is pointwise and uniformly convergent to $0$. But what is this "limit-function" supposed to look like? Can I just say $f(x)= 0$ ? The derivative would be $0$ too… But whats the point in this task then?
- Check if ${f_n'}$ is pointwise/uniformly convergent. Specify the "limit-function" if there is one.
So $f_n'= \frac{1-n^2x^2}{(1+n^2x^2)^2}$
The problem is, that pointwise and uniform convergence are totally confusing me. To see if the sequence is pointwise converging, we show that $\lim\limits_{n\to\infty} f_n'(0)=1,\lim\limits_{n\to\infty} f_n'(1)=0,\lim\limits_{n\to\infty} f_n'(-1)=0$ and thus is converging to $f(x)=0$ for [$-1,0$) ($0,1$] and $f(x)=1$ for $x=0$ Is this correct? Unfortunately I am not able to show if the sequence is uniformly converging. The only thing I have is that $|f_n'|\leq \frac{1-n^2x^2}{4n^2x^2}=\frac{1}{4n^2x^2}-\frac{1}{4}$.
Is there anyone who could give me an advice?
(Also this is not a duplicate, since the focus of this question lays on the "limit-function" and the derivative and I couldn't find anything that could solve my problem)
Best Answer
We have that $\displaystyle f_n'(x) = \frac{1 - n^2x^2}{(1+ n^2x^2)^2}$ converges pointwise to $\displaystyle g(x) = \begin{cases} 1, & x= 0 \\0, & x \in [-1,1] \setminus \{0\}\end{cases}$
Thus,
$$|f'_n(x) - g(x)| = \begin{cases} 0, & x= 0 \\ \left|\frac{1 - n^2x^2}{(1+ n^2x^2)^2}\right|,& x \in [-1,1] \setminus \{0\}\end{cases}$$
For uniform convergence to hold, we must have $\sup_{x \in [-1,1]}|f_n'(x) - g(x)| \to 0$ as $n \to \infty$
However,
$$\sup_{x \in [-1,1]}|f_n'(x) - g(x)| \geqslant \left|f_n'\left( \frac{1}{\sqrt{2}n}\right) - g\left( \frac{1}{\sqrt{2}n}\right)\right| = \frac{1 - n^2\cdot \left(\frac{1}{\sqrt{2}n} \right)^2 }{\left(1 + n^2\cdot \left(\frac{1}{\sqrt{2}n} \right)^2\right)^2 } \\=\frac{1- \frac{1}{2}}{\left(1 + \frac{1}{2}\right)^2} = \frac{2}{9} \underset{n \to \infty}{\not\to0}$$