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!
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 $I$.