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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 $\{f_n\}\overset{unif.}{\not\to} 0$.

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    $\begingroup$ I've taken the liberty of adding the solution-verification and sequence-of-function tags to your question. $\endgroup$ Dec 19, 2020 at 2:20

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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.

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  • $\begingroup$ Could you explain why the sequence doesn’t converge at $x=0$? $\endgroup$
    – user853982
    Dec 19, 2020 at 2:32
  • $\begingroup$ @D.Math Sorry about that -- in some places you used $\chi_{(0,1/n]}$ and in some places you used $\chi_{[0,1/n]}$, and I got confused about which was which. I've updated the answer. $\endgroup$ Dec 19, 2020 at 2:34
  • $\begingroup$ Oh I see, i’ll fix that. So, for the $f_n$ I defined it does converge at $x=0$? $\endgroup$
    – user853982
    Dec 19, 2020 at 2:37
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    $\begingroup$ Yes -- $f_n(0)=0$ for all $n$ the way you defined it. $\endgroup$ Dec 19, 2020 at 3:25

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