# A sequence that converges pointwise but not uniformly.

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

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

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.

• Could you explain why the sequence doesn’t converge at $x=0$?
– user853982
Dec 19, 2020 at 2:32
• @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. Dec 19, 2020 at 2:34
• Oh I see, i’ll fix that. So, for the $f_n$ I defined it does converge at $x=0$?
– user853982
Dec 19, 2020 at 2:37
• Yes -- $f_n(0)=0$ for all $n$ the way you defined it. Dec 19, 2020 at 3:25