Find the limit function if this exist

Question

I begin study the uniformly and pointwise convergence of functions and I have this sequence function:

$$X=\mathbb{R}\hspace{1cm}f_{n}(x) = \left\{\begin{array}{b}\ n& \text{if }-n\leq x\leq n\\ 0&\text{ if } |x|>n \end{array}\right.$$

I need to find the limit function but I think in this case doesn't exist because throughout the sequence for some interval $[-k,k]$ this convergence to a constant function $k$ ($k \in \mathbb{N}$) but I can take some $x$ out of this interval and I can find some $\varepsilon_{0}>0$ such that: \begin{eqnarray*} |f_{n}(x)-f(x)|&=&K\\ &>& \varepsilon_{0} \end{eqnarray*}
But I'm not sure this is correct, I am confusion. Can you give me some advice or hint to resolve this? Thank you.

Answer 1

Your conclusion is correct, but your statement that on $[-k, k]$ the sequence converges to a constant function is not since $f_k(x)$ grows without bound as $k \to +\infty$ for every $x \in \mathbb{R}$. In fact, we can use this unbounded growth to prove the lack of convergence.

Assume $f_k$ tends to $f$ pointwise, i.e. $f_k(x) \to f(x)$ for every $x \in \mathbb{R}$. Fix $x$ and let $N$ be the smallest integer greater than $\max(|x|, f(x))$. Note that $f(x) \leq N$ and $f_k(x) = k$ for every $k > N$. Consequently,

$$ |f_k(x) - f(x)| \geq |k - N| > 1 $$

so $f_n$ does not converge to $f$ pointwise.

Since $f_n$ does not converge pointwise, it also does not converge uniformly.

Answer 2

Fix $x\in\Bbb R$ and take $n_0=\lfloor |x|\rfloor +1$ (the floor of $|x|$ plus 1).

Then for any $n\ge n_0$ is clear that $|x|<n$, so $f_n(x)=n$.

This implies that $$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty} n=+\infty$$

So, the sequence is pointwise divergent.