Decide if $\{f_n(x) : n\in\mathbb{N}\}$ converges to $f$ uniformly or not.

Question

Can someone please help me with the problem below? I am not confident in my solution. I am working off Munkrees' Topology book and problems in the book.

For each $n\in\mathbb{N}$, let $f_n: [0,1] \to [0,1]$ be defined by $f_n(x) = 0, x > \frac{1}{n}$ and $f_n(x) = 1-nx$ if $0\le x\le\frac{1}{n}$. The collection $\{f_n(x) : n\in\mathbb{N}\}$ converges to a point, call it $f(x)$ for each $x\in [0,1].$ Decide if $\{f_n(x) : n\in\mathbb{N}\}$ converges to $f$ uniformly or not.

$\textbf{Solution:}$ Observe, $f_n(0) = 1$ for all $n$. For $0< x \le 1$ by Archimedean property of real numbers, there exist $N$, a fixed natural number, such that $\frac{1}{N} < x$. So for all $n\ge N, \frac{1}{n} \le \frac{1}{N}.$ Hence, for all $n\ge N, \frac{1}{n} < x$, and $f_n(x)=0$, by definition. We observe this because for $0 < x \le 1$, after finitely many non-zero $f_n(x)$ all other $f_n(x)=0$. So, $f(0)=1$ and $f(x)=0, 0 < x\le 1$ as $x=0$, it is the constant sequence $1$ and for $0<x\le 1$, after finitely many non-zero terms, the sequence is eventually constant sequence $0$.

We will show $f_n(x)$ does not converge to $f(x)$ uniformly.

Take the $\sup|f_n(x)-f(x)| = 1$ where the supremum runs over $0\le x \le 1$. Since, $1$ does not tend to zero. Hence, we are done.

Furthermore, there is a theorem that if the sequences of continuous functions converge uniformly then the limit of the function is continuous. Here, $f_n(x)$ are continuous throughout and $f(x)$ is not continuous at $x=0$. So, they cannot converge uniformly.

Answer 1

The pointwise limit, as you correctly showed: for each $x \in [0,1]$ $\lim_n f_n(x) =f(x)$ where $$f(x)=\begin{cases} 1 & x=0\\0 & 0<x \le 1\end{cases}$$

Now, if Munkres treated the theorem that a uniform limit of continuous functions is continuous, we can already conclude that $(f_n)_n$ does not converge uniformly: the pointwise limit is the only candidate for a uniform limit. But this can be shown directly by computing $\sup_x |f_n(x)-f(x)| = 1 $ as you did. But that fact does need justification! Don't just claim it..

Another way is to apply another theorem and compute $f_n(\frac{1}{2n})= 1-n\frac{1}{2n} = \frac12 \not\to 1=f(0)$ (based on the theorem:

$f_n \to f$ uniformly and $x_n \to x$ in the domain, then $f_n(x_n) \to f(x)$

which you might or might not know (the proof is not hard from the definitions, applying the triangle inequality).