Recently, I am learning functional analysis written by Philippe G. Ciarlet. In page 55, he gives a definition of converge locally uniformly and an example. I have some ideas about how to analysis the example, but I don't solve it completely. I will list the definition of converge locally uniformly, then talk about my problems.
Let $X$ be a topological space and $Y$ be a normed vector space. Then a sequence $f_n:X\to Y$ is said to converge locally uniformly to a mapping $f:X\to Y$ as $n\to\infty$ if, given any $x_0\in X$, there exists a neighborhood $V(x_0)$ of $x_0$ such that $$\lim\limits_{n\to\infty}\sup\limits_{x\in V(x_0)}||f_n(x)-f(x)||=0.$$
Example
$f_n:x\in(0,+\infty)\to\displaystyle\frac{1}{x}+\max(0,x-n)$, then $f_n$ converges locally uniformly to $f:x\in(0,+\infty)\to\displaystyle\frac{1}{x}$.
Here are my ideas. When $x_0<n$, we can choose $V(x_0)$, such that $V(x_0)\subset(0,n)$, so for any $x_0\in V(x_0), f_n(x)-f(x)=0$, i.e. When$x_0<n$, we have $f_n(x_0)$ converge locally uniformly to $f(x_0)$. But when $x_0>n$, for any $V(x_0)$, it can be written$V(x_0)=(x_0-\epsilon,x_0+\epsilon)$, then $$\sup_{x\in V(x_0)}||f_n(x)-f(x)||=\sup_{x\in V(x_0)}(x_0+\epsilon-n),$$ $$\lim_{n\to\infty}\sup_{x\in V(x_0)}||f_n(x)-f(x)||=-\infty.$$
So it is not converge locally uniformly when $x_0>n$, I don't know which step I went wrong.
The variable $n$ is a dummy variable in some sense; the neighborhood, and the point $x_0$, do not depend on it. ($x_0$ may be bigger than $n$ when $n = 5$, but how about when $n = 500$? And if it's bigger than 500, how about when $n = 5000$?) In fact, $n$ should depend on $x_0$ and $V(x_0)$, not the other way around.
First you pick an $x_0$, and then you pick a neighborhood $V(x_0)$. Then you hope that, as $n \to \infty$, $\lim_{n \to \infty} \sup_{x \in V(x_0)} |f_n(x) - f(x)| = 0$. (And by "hope", here, I mean "show".)
One observation that may come in handy: for any $x_0$, $x_0$ is eventually less than $n$ as $n \to \infty$.