Prove that $\{f_n\}$ converges pointwise but not uniformly on $\Bbb R$, where $f_n(x)=\frac{1-|x|^n}{1+|x|^n}$ .

Question

For each $n\in \Bbb N$ and $x\in \Bbb R$, define $$\begin{align} f_n(x)=\frac{1-|x|^n}{1+|x|^n}_. \end{align} $$ Prove that $\{f_n\}$ converges pointwise but not uniformly on $\Bbb R$.

For $x\in (-1,1)$, $f_n(x)\to 1$, and for $x>1$ or $x< -1$, $ f_n(x)=\frac{1-|x|^n}{1+|x|^n} \to -1$. Define the limit function $\begin{align} f(x)=\begin{cases} 1, & \text{if $x\in (-1,1)$ } \\ -1, & \text{if $x>1$ or $x< -1$} \end{cases}_. \end{align}$ Then $f_n \to f$ pointwise on $\Bbb R$.

Let $\varepsilon_n=\sup\{\left|f(x)-f_n(x) \right|:x\in\Bbb R \}$, how can I get a contradiction by showing that $\varepsilon_n$ do not converge to $0$?

Answer 1

Your function $f$ is not complete: for $|x|=1$ we have $f_n(x) = 0 \to 0,$ hence $f(x)=0.$

The limit function $f$ is not continuous, but all $f_n$ are continuous ! Conclusion ?

Answer 2

A simple argument

As all $f_n$ are continuous, if the converge to $f$ was uniform, $f$ would have to be continuous. As it is not the case, the convergence is not uniform.