I'm reading my notes about uniform convergence, and this question came to my mind:
Let $g:\mathbb{R} \to (0,+\infty)$ be a continuous function, for example $g(x)=\displaystyle\frac{1}{e^{x}}$. Let
$$f_n(x)=\frac{ng(x)}{1+ng(x)}.$$
We have that $f_n$ converges pointwise to $1$ in $\mathbb{R}$.
It's easy to see that $f_n$ converges uniformly to $1$ in $[a,b]$, because $g(x)$ has a minimum.
But it converges uniformly in $[0,+\infty)$?
I think that it doesn't converge uniformly, because
$$\sup_{x\in[0,+\infty)}\mid\frac{ng(x)}{1+ng(x)}-1\mid=\sup_{x\in[0,+\infty)}\frac{1}{1+ng(x)},$$
and if we look at $g(x)=\displaystyle\frac{1}{e^{x}}$, we see that it doesn't have a minimum, so the supremum never tends to $0$, and it's not always true that $f_n(x)$ is uniformly convergent in $[0,+\infty)$ for every $g(x)$ continuous and with positive images.
Is this true? Thanks!
You are right. In fact let $g$ be any (positive) continuos function with $\inf_{x \in (0, +\infty)} g(x) = 0$. for a fixed $n$, it's easy to see that
$$\sup_{x \in [0, \infty)} \frac 1{1 + ng(x)} = 1$$ which does not tend to $0$.
If on the other hand $g(x)$ stays "away" from $0$, that is $\inf g(x) = l > 0$, then $$\sup_{x \in [0, \infty)} \frac 1{1 + ng(x)} = \frac1{1 + nl} \to 0$$
so in this case we have uniform convergence.