# Does pointwise convergence implies uniform convergence when the limit is continous?

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.

• This should be a counter example: math.stackexchange.com/questions/1606121/…. Feb 27, 2016 at 12:39
• 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/… Feb 27, 2016 at 15:28
• 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. Feb 27, 2016 at 15:56

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).
• Hello, thanks for the quick answer. What is the case if $f(\infty)=a>0$? Feb 27, 2016 at 12:53
• @mathcounterexamples.net Just add some continuous function $g(x)$ (like the constant function $1$) to all $f_n(x)$'s. Feb 27, 2016 at 13:00