I am reading through Roydens Real Analysis text and I have not studied, at least in depth, sequences of functions. Thus, I have taken a slight aside from reading the text in order to look at sequences of functions. I am curious if the following is an example of a sequence of functions on $[0,1]$ that converges pointwise but not uniformly. Moreover, if the follow up argument provides a proof of this claim.
Define the sequence of functions $\{f_n\}$ as:
$$f_n(x):=n\chi_{(0,1/n]}.$$
Or written differently:
$$
f_n:=\begin{cases}
n & \text{ if } 0< x \leq 1/n \\
0 & \text{ otherwise } \\
\end{cases}
$$
Naturally, we have the following for a fixed $x$:
$$\lim\limits_{n\to\infty}f_n = \lim\limits_{n\to\infty}n\chi_{(0,1/n]} = \lim\limits_{n\to\infty}
\begin{cases}
n & \text{ if } 0 < x \leq 1/n \\
0 & \text{ otherwise } \\
\end{cases} =
\begin{cases}
\infty & \text{ if } 0 < x \leq 0 \\
0 & \text{ otherwise } \\
\end{cases} =
0.$$
Therefore, we can conclude that $\{f_n\}\overset{point.}{\to} 0$. However, we claim that
$\{f_n\}\overset{unif.}{\not\to} 0$. To see this, define the following:
$$\Delta_n:=\sup\limits_{x\in[0,1]}\left|f_n(x)-f(x)\right| = \sup\limits_{x\in[0,1]}\left|n\chi_{[0,1/n]} – 0\right| = \sup\limits_{x\in[0,1]}\left|n\chi_{[0,1/n]}\right| =
\sup\limits_{x\in[0,1]}n\chi_{[0,1/n]} =
n.$$
Since
$$\lim\limits_{n\to\infty}\Delta_n = \lim\limits_{n\to\infty}n\neq 0,$$
we can conclude that $\{f_n\}\overset{unif.}{\not\to} 0$.
Best Answer
This is essentially correct. In some places you use $\chi_{[0,1/n]}$ instead of $\chi_{(0,1/n]}$ -- for these functions, the sequence doesn't converge at $x=0$.
You can make your functions converge pointwise on all of $[0,1]$ bounded by removing the factor of $n$: if $f_n=\chi_{(0,1/n]}$, then $f_n(x)$ converges to $0$ for all $x$.
You can also make the functions continuous if you wish by replacing the drop from $n$ to $0$ (or from $1$ to $0$ as in the previous paragraph) a very short line segment, and moving your interval to somewhere in the middle of $[0,1]$, so that you can do this for both endpoints.