As I was going over the definitions of pointwise and uniform convergence I came to the following problem: since the canonical example for continuous functions on $[0,1)$ which are pointwise but bot uniform convergent(wrt the constant function $f=0$) is sequence of functions $f_n(x) = x^n$ I ask myself is there such sequence of functions for the interval $[0,1]$. So far I couldn't think find any example and I am stating to believe that the answer might be negative. So what do think, is there such sequence and if not can it be proven that such sequence does not exist?

  • $\begingroup$ The same sequence is pointwise but not uniformly convergent on $[0,1]$. $\endgroup$
    – Lukas Betz
    May 13, 2015 at 20:59
  • $\begingroup$ @LeBtz maybe forgot to mention that we want convergence to the constant function $f=0$ $\endgroup$
    – sve
    May 13, 2015 at 21:01
  • $\begingroup$ I guess you might want to check Dini's theorem $\endgroup$ May 13, 2015 at 21:05

1 Answer 1


Certainly. take for example $f_n(x) = nxe^{-nx}$ which converges pointwise but not uniformly to $0$ on $[0,1]$. (And as pointed out in the comment, your own example works as well.)

  • 1
    $\begingroup$ Mine example does not work when $x=1$ but yours works brilliantly since for every $n$ we have that $f_n(\frac{1}{n}) = e^{-1}$ which breaks the uniform convergence requirement. How did you come up with this function? $\endgroup$
    – sve
    May 13, 2015 at 21:22
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    $\begingroup$ @lnwvr Start with any function $g(x)$ such that $g(0) = 0$ and $\lim_{x\to\infty} g(x) = 0$ and put $f_n(x) = g(nx)$. (Think graphically: how does the graph of $f_n$ compare to the graph of $g$?) If $g$ is not the zero function, you get pointwise but not uniform convergence to $0$ on any finite interval $[0,a]$. $\endgroup$
    – mrf
    May 13, 2015 at 21:24
  • $\begingroup$ wow! @ i speaketh out of my naivety. @sve, you have amazingly demonstrated the existence of an 'x=1/n' for each 'n' such that |f_n(x)-f(x)|=|(e^-1)-(0)|=e^-1, which will not work for any $\varepsilon < e^-1$. Super cool. $\endgroup$
    – Krishan
    Nov 22, 2020 at 7:40

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