Finding "limit-function" of $f_n=\frac{x}{1+n^2x^2}$, Pointwise/Uniform convergence of ${f_n'}$

Question

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)

Answer 1

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,

$$\sup_{x \in [-1,1]}|f_n'(x) - g(x)| \geqslant \left|f_n'\left( \frac{1}{\sqrt{2}n}\right) - g\left( \frac{1}{\sqrt{2}n}\right)\right| = \frac{1 - n^2\cdot \left(\frac{1}{\sqrt{2}n} \right)^2 }{\left(1 + n^2\cdot \left(\frac{1}{\sqrt{2}n} \right)^2\right)^2 } \\=\frac{1- \frac{1}{2}}{\left(1 + \frac{1}{2}\right)^2} = \frac{2}{9} \underset{n \to \infty}{\not\to0}$$

Answer 2

Assume $f_n' \to g$ uniformly on $[-1,1]$.

Since $f_n'(0)=1$ for every $n \in \mathbb N$, we have $g(0)=\lim_{n \to \infty} f_n'(0)=1$. Moreover, $g(x)=\lim_{n \to \infty} f_n'(x)=0$ for every $x \in [-1,1]\setminus \{0\}$.

Consider $\varepsilon=\frac{1}{2}$. By our assumption there is $N \in \mathbb N$ so that $|f_N'(x)-g(x)|<\frac{1}{2}$ for all $x \in [-1,1]$.

Since $f_N'$ is continuous at $0$ (being a rational function with no singularity at $0$) there is $\delta>0$ so that $|f_N'(x)-1|<\frac{1}{2}$ whenever $|x|<\delta$. Now notice that for $x \in (-\delta, \delta)\setminus \{0\}$ we have \begin{aligned} \left|f_N'(x)-g(x)\right|&=\left|\left(f_N'(x)-1\right)-\left(g(x)-1\right)\right| \\&\geq \left|\left|f_N'(x)-1\right|-\left|g(x)-1\right|\right| \quad (\text{Reverse Triangle Inequality}) \\&=\left|\left|f_N'(x)-1\right|-1\right| \\&=1-\left|f_N'(x)-1\right|>1-\frac{1}{2}=\frac{1}{2}. \end{aligned} This is a contradiction. Therefore $\{f_n'\}$ does not converge uniformly on $[-1,1]$.

Answer 3

$f_n'(x)=\frac{1-n^2x^2}{(1+n^2x^2)^2}$

$f_n'(0)=1\implies \lim_{n\to \infty}f_n'(0)=1$

And for $x\ne0, f_n'(x)= \frac{1-n^2x^2}{(1+n^2x^2)^2}\to 0$. So the limit function will be:

$g(x)=\begin{cases}0; x\ne 0\\1; x=0\end{cases}$

If the convergence $f_n'(x)\to g(x)$ were uniform then by continuity of $f_n'$'s, $g(x)$ should have been continuous but since this is not the case, it follows that the convergence is not uniform.