# Uniform convergence on bounded interval

Consider the function

$$f_n(x)=\frac1{nx+1}$$ on $x\in]0,1]$.

I'm trying to find if $f_n\to f$ uniformly on $]0,1]$, but I don't feel like I'm getting the definitions correctly. What I've attempted to do is find out the following limit:

$$\lim_{n\to\infty}\sup_{0<x\leq1}\left\vert\frac1{nx+1}\right\vert$$

Then I think the supremum of $f_n$ on the interval is $1$, and so the limit is $1$. Therefore, no uniform convergence.

• @User is correct, I had the same thought but then I realized the question was about uniform convergence in the half-open interval. ([ means ( in some European countries) Commented May 15, 2018 at 23:16
• Well, on $(0,1]$ we have $f_n\to 0$. But, for $\epsilon=\frac12$, we have for all $N$ there exists a number $n>N$ and an $x\in (0,1]$ (take $x=1/n$) such that $$\left|f_n(x)-0\right|]\ge\frac12$$ Commented May 15, 2018 at 23:25

What you can easily claim is that the supremum is at least $1$, because
$$\sup_{0<x\le 1} \frac1{nx + 1} \ge \lim_{x \to 0} \frac1{nx + 1} = 1$$
so yes, the limit is at least $\lim_{n\to \infty} (1) = 1$, so it is non-zero and convergence is not uniform.
• I have attempted to solve $|f_n(x)-f(x)|<\varepsilon$ and got $\frac{1-\varepsilon}{x\varepsilon}<n$. If for some other pointwise convergent function we happened to have $x$ in the numberator we could somehow establish a lower bound for how big $n \geq N$ has to be, correct? (because we are working in a bounded interval.) Commented May 15, 2018 at 23:19