I have recently come across pointwise/uniformly convergent sequences of functions, and I am hoping if someone could give some examples of certain sequences of functions so that I could understand the concept better. Thanks!

• Pointwise convergent sequences that do / do not preserve continuity,

• Pointwise convergent sequences that do / do not preserve integrals.

It seems that $f_n(x) = x^n$ is an example that do not preserve continuity?

  • $\begingroup$ Hi and welcome to Math.SE. Your question is not clear: a sequence of complex numbers is always pointwise converging and uniform convergence has not any meaning for such an object. Are you asking about sequences of functions? If so, please edit your question and take a little care of it. $\endgroup$ Dec 12, 2018 at 15:18
  • $\begingroup$ @DanieleTampieri Thanks, I have made it more clear. $\endgroup$
    – Valia
    Dec 12, 2018 at 15:22

1 Answer 1


Here are two useful (non)examples and one somewhat trivial well-behaved example.

Example 1. Take $f_n(x)=\frac{1}{1+x^n}$ for $x\in [0,1]$. By a direct calculation, we see $$ \lim_{n\to\infty} f_n(x)= \begin{cases} 1&x\in [0,1)\\ \frac{1}{2}&x=1. \end{cases}$$ Here, the pointwise limit has a discontinuity. You should try to prove why this is not a uniformly convergent family.

Example 2. Define the characteristic function $$ \chi_{[0,\frac{1}{n}]}(x)= \begin{cases} 1&x\in [0,1/n]\\ 0& x\not\in [0,1/n]. \end{cases}$$ Next. set $f_n(x)=n\chi_{[0,\frac{1}{n}]}$. Then for any $n$, we have $$ \int_0^1 f_n(x)dx=1.$$ However, $\lim_{n\to\infty} f_n(x)=f(x)\equiv 0.$ In particular, we do not have uniform convergence. As a consequence, we can see that $$ \int_0^1\lim_{n\to\infty} f_n(x)dx=\int_0^10\cdot dx=0.$$ So, the limit does not preserve the integral in this case.

Example 3. An easy example of uniform convergence is something like $f_n(x)=\frac{n}{n+1} x$ on $[0,1]$ which is continuous. Clearly, $\lim_{n\to\infty} f_n(x)= f(x)=x$ on $[0,1]$. Furthermore, $$ \lim_{n\to\infty}\int_0^1f_n(x)dx=\lim_{n\to\infty}\int_0^1\frac{n}{n+1}x dx=\lim_{n\to\infty} \frac{n}{n+1}\int_0^1 xdx=\int_0^1f(x)dx.$$

  • $\begingroup$ I'm wondering about your second example. I'm struggling to see how $f(x)=0$ when $n\to \infty$. If $x\neq 0$, then yes $f(x)=0$. But If $x=0,$ then $f(x)$ is infinity, right? $\endgroup$ Aug 21, 2022 at 19:35

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