Differentiability and Uniform convergence on unbounded intervals 
Let $I$ be a bounded interval of $\mathbb{R}$, and let $\{f_{n}:I\to\mathbb{R}\}$ be a sequence of differentiable functions. Suppose that a sequence $\{f_{n}(x_{0})\}$ converges for some $x_{0}\in I$ and that $\{f_{n}'\}$ converges uniformly to some function $g:I\to\mathbb{R}$. Then, $\{f_{n}\}$ converges uniformly to $f:I\to\mathbb{R}$ so that $f'\equiv g$ on $I$.

It is well-known theorem.
I've been using it when it is hard to show directly that the convergence of $\{f_{n}\}$ is uniform.
However, when I solved the problem, in many cases, the interval is $\bf{unbounded}$, for example $(0,\infty),~\mathbb{R},$ etc.
I know that the above theorem is fail if $I$ is not bounded, and the supremum norm method.
Is there a generalized above theorem(I means the case of $I$ is unbounded interval) or additional hypothesis?
I have never seen the use of the above theorem to proving the uniform convergence on unbounded intervals.
Give some examples or reference any kinds! Thank you!
 A: The theorem doesn't necessarily hold when $I$ is unbounded. Here's an example
We take the sequence $\{f_{n}\}_{n\in\mathbb{N}}$, where $f_{n}(x)=\dfrac{x}{n}$, for every $x\in \mathbb{R}$. The derivative of $f_{n}$ is $$f_{n}^{'}(x)=\dfrac{1}{n} \quad \forall x\in \mathbb{R} $$ The sequence $\{f_{n}^{'}\}_{n\in\mathbb{N}}$ is convergent to $g:\mathbb{R}\longrightarrow\mathbb{R}$, given by $g(x)=0$. And that convergence is uniform in $\mathbb{R}$, since $$ \lim_{n\to\infty}\Big[\sup_{x\in\mathbb{R}}|f_{n}^{'}(x)-g(x)|\Big]=\lim_{n\to\infty}\Bigg|\dfrac{1}{n}-0 \Bigg|=0 $$
And, even though $\{f_{n}\}_{n\in\mathbb{N}}$ is pointwise convergent to $f(x)=0$, it does not converge uniformly, since for every $n$, the function $|f_{n}(x)-f(x)|=\dfrac{|x|}{n}$ is unbounded in $\mathbb{R}$.
Now, even though the convergence is not uniform in this case, it's still true that $f'=g$. That's because the second part of the theorem is still true. 
The reason is the following: if $I$ is open, whether it's unbounded or not, and $\{f_{n}\}_{n\in\mathbb{N}}$ converges pointwise, for every $x_{0}\in I$, there exists an open interval $(x_{0}-\delta,x_{0}+\delta)$ contained in $I$, and thus we can apply that theorem in $(x_{0}-\delta,x_{0}+\delta)$: the sequence converges uniformly in that interval and $f'=g$ in the interval, so in particular $f'(x_{0})=g(x_{0})$. Since the choice of $x_{0}$ is arbitrary, we deduce that $f'=g$ in $I$.
