# Uniform convergence of indicators

Suppose $f_n(x)$ is a sequence of real-valued functions converging uniformly to a function $f(x)$ and let $\{\epsilon_n\}_{n \in \mathbb{N}}$ be a sequence of positive real numbers converging to $0$.
Is it necessarily true that the sequence of (extended real-valued) functions $$g_n(x)\triangleq \frac{1}{\epsilon_n}I_{\{ f_x(x)>\epsilon_n \}}$$ converges uniformly to the (extended real-valued) function \begin{aligned} g(x)\triangleq & \begin{cases} \infty & \mbox{if} & f(x)>0\\ 0 & \mbox{if} & f(x)=0? \end{cases} \end{aligned} If not what additional conditions must I impose for this to be true?

## 1 Answer

No. Consider $\epsilon_n = 2^{-n}$ and $f_n(x) = \frac{1}{n}$. Then $g_n(x) = 2^n$ everywhere, but $g(x) = 0$ everywhere. You need $\epsilon_n$ to converge to $0$ faster than $f_n$ where $f(x)=0$. I don't know of a simple sufficient condition.

• So if I choose $\epsilon_n$ to be of higher order of convergence than $f_n$ then I'll be ok? – AIM_BLB Sep 5 '17 at 17:55