Real Analysis - Uniform Convergence of $f_n$

Question

I am given that:

For $n \in \mathbb{N}$, define $f_n: \mathbb{R} \to \mathbb{R}$ by

$$f_n(x)=\frac{x^{4n}}{4+x^{4n}}.$$

I need to determine whether the sequence $(f_n)$ converges uniformly on $\mathbb{R}$.

This is what I have done:

\begin{align} \lim_{n \to \infty}f_n(x)&=\lim_{n \to \infty}\frac{x^{4n}}{4+x^{4n}}= \begin{cases} 0, & \text{if}\ x \in (-1,1) \\ \frac15, & \text{if}\ x \in \{-1,1\} \\ 1, & \text{if}\ x \in \mathbb{R}\setminus[-1,1] \end{cases} = f(x) \end{align}

Now I am struggling to show whether or not $f_n$ converges uniformly to $f$ over $\mathbb{R}$.

Answer 1

Well, $f_n$ are continuous functions. If $(f_n(x))$ is uniformly convergent , then $\lim_{n \to \infty}f_n(x)$ is continuous, which is false.

So, our sequence is not uniformly convergent over $\mathbb{R}$.

Answer 2

A useful theorem states that if $f_n \rightarrow f$ uniformly, then $f$ must be continuous. Since your pointwise limit function is not continuous, the convergence cannot be uniform.

If you don't want to use the theorem, you can check whether $\mathrm{sup}_{x \in \mathbb{R}} |f_n(x) - f(x)| \rightarrow 0$.

Since

$$ \mathrm{sup}_{x \in \mathbb{R}} |f_n(x) - f(x)| \geq \mathrm{sup}_{|x| < 1} |f_n(x) - f(x)| = \mathrm{max}_{|x| \leq 1} \frac{x^{4n}}{4 + x^{4n}} = \frac{1}{5} $$

we see that the convergence cannot be uniform (even on the interval $(-1,1)$ and surely not on $\mathbb{R}$).