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

Question

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.

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).