Uniform Convergence of functions in a ray, using Dini`s Theorem Let $f_n(x) =\sqrt[n] {1+x^n}$, defined for $x\geq0.
$
pointwise; $f_n(x) \rightarrow f(x)=\begin{cases}
1,x\in[0,1)\\
x  , x\geq 1
\end{cases}
$
I was wondering about uniform convergence.
$f_n$ are monotone (relative to $n$) for any $x\in[a,b]$ for any $a,b\geq0$, (Decreasing for $x\leq1$ and increasing for $x>1)$, and $f$ is continous, so we can use Dini`s theorem to show uniform convergence,in the closed interval $[a,b]$. However, does this mean $f_n$ converge uniformly in $(0,\infty$)?
The example above is just an excercise I stumbled upon that lead me to this question, I'd much rather have a general answer if we know that $f_n$ are monotone (relative to $n$) for any $x\in[a,b]$ for any $a,b\geq0$, does this imply $f_n$ converge uniformly in $(0,\infty$)?
 A: Dini's theorem states that if $(g_n)_n$ is a monotone sequence of continuous functions $g_n : [a,b] \to \mathbb{R}$ converging pointwise to $g : [a,b] \to \mathbb{R}$, then the convergence is uniform.
In your example using Dini you can only conclude that $f_n|_{[0,1]} \to 1$ uniformly on $[0,1]$ and that $f_n|_{[a,b]} \to x$ uniformly on all segments $[a,b] \subseteq [1, +\infty)$.
However, the convergence indeed is uniform on $[0, +\infty)$. To show this, note that for $x \ge 1$ we have
$$1 = (1+x^n) - x^n = \left(\sqrt[n]{1+x^n}\right)^n - x^n = \left(\sqrt[n]{1+x^n} - x\right)\sum_{k=0}^{n-1} \left(\sqrt[n]{1+x^n}\right)^{n-k}x^k \ge n \left(\sqrt[n]{1+x^n} - x\right)$$
so $\sqrt[n]{1+x^n} - x \le \frac1n \xrightarrow{n\to\infty} 0$ uniformly on $[1, +\infty)$. This together with the uniform convergence $f_n|_{[0,1]} \to 1$ gives $f_n \to f$ uniformly on $[0, +\infty)$.
In general, if $g_n \to g$ pointwise on $(0, +\infty)$ and uniformly on all segments $[a,b] \subseteq (0, +\infty)$, it still doesn't follow that the convergence is uniform on $(0, +\infty)$.
An example is the sequence of partials sums of the exponential function $g_n(x) = \sum_{k=1}^n \frac{x^k}{k!}$. Then $g_n \to \exp$ uniformly on all segments $[a,b] \subseteq (0, +\infty)$:
$$\exp(x) - g_n(x) = \sum_{k=n+1}^\infty \frac{x^k}{k!} \le \sum_{k=n+1}^\infty \frac{b^k}{k!} \xrightarrow{n\to\infty} 0$$
 but the convergence is not uniform on $(0, +\infty)$. I discussed this example here.
