Suppose we have a series of functions $f_n: \mathbb R \rightarrow [0,1]$ and continous function $f: \mathbb R \rightarrow [0,1]$. Suppose that $f_n \rightarrow f$ pointwise as $n \rightarrow \infty$. Is it true, that $f_n$ converges uniformly also? What is the case if all $f_n$ and $f$ are monotone functions?

Edit1: Consider the case when $f_n$ and $f$ are distribution functions. All are monoton, continous functions with $$\lim_{x\to -\infty}f(x)=0$$ and $$\lim_{x\to\infty}f(x)=1.$$

There are numerous question on the site already, the most relevant is this: Does pointwise convergence against a continuous function imply uniform convergence?

In the marked answer, there are two counterexamples, but I think both example series of functions converge to a $g(x)=\delta(x)$, which is not continous. Am I right? If yes, how the original statement could be proved?

I am also aware of Dini's Theorem, but that applies only to function on closed intervals.

  • $\begingroup$ This should be a counter example: math.stackexchange.com/questions/1606121/…. $\endgroup$
    – Martin R
    Feb 27, 2016 at 12:39
  • $\begingroup$ If they are distribution functions, then, yes, they converge uniformly under the stated assumptions. In other words: convergence in distribution of $X_n$ to $X$, where $X$ has continuous cdf $F$ implies that $F_n$ converge uniformly to $F$. Here is the relevant question/answer: math.stackexchange.com/questions/1670030/… $\endgroup$
    – Jimmy R.
    Feb 27, 2016 at 15:28
  • $\begingroup$ There are, in essence, two questions here. The first is if $f_n\rightarrow f$ pointwise on $\mathbb{R}$, with all functions continuous, is the convergence uniform. This statement is false as the link above and the example below illustrate (the example below converges to the zero function and not a delta). If you add the conditions that $f_n$ and $f$ are cdf's, then the statement is true, as the other link above and the link below show. $\endgroup$ Feb 27, 2016 at 15:56

1 Answer 1


Consider a index function $f_n(x)=I_{(n,n+1)}(x)$. This converges pointwise to the zero function, but the convergence is not uniform. You can replace $f_n(x)$ with a continuous function (a bump function) and the same idea works. If you want a monotone function, use $f_n(x)=I_{(n,\infty)}(x)$ (or a continuous version of this).

  • $\begingroup$ Hello, thanks for the quick answer. What is the case if $f(\infty)=a>0$? $\endgroup$ Feb 27, 2016 at 12:53
  • $\begingroup$ @HodossySzabolcs In that case the convergence is uniform. $\endgroup$ Feb 27, 2016 at 13:00
  • $\begingroup$ @mathcounterexamples.net Just add some continuous function $g(x)$ (like the constant function $1$) to all $f_n(x)$'s. $\endgroup$ Feb 27, 2016 at 13:00
  • $\begingroup$ @mathcounterexamples.net: How could you prove it? Also to tell the truth, I would like to construct a proof for distribution functions, I just think that it is a purely related to analysis and not probability theory. $\endgroup$ Feb 27, 2016 at 13:06
  • $\begingroup$ I have this reference on French Wikipedia fr.m.wikipedia.org/wiki/Théorèmes_de_Dini. Unfortunately, the English version is less detailed. $\endgroup$ Feb 27, 2016 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.