For each $n\in \mathbb N$, let $f_n:\mathbb R\to \mathbb R$ be $C^1$.
Suppose
・$\forall x\in \mathbb R$ ; $\displaystyle\lim_{n\to \infty}f_n(x)= f(x)$
・$f'_n \to g$ uniformly.
Then, does $\bigg[$$f$ is $C^1$ on $\mathbb R$ and $f'=g \bigg]$ hold ?
I'm not sure whether the proof below works.
Fix $x\in \mathbb R$.
I have $f_n(x)=f_n(0)+\int_0^x f'_n (y) dy$.
Letting $n\to \infty$, I get $f(x)=f(0)+\int_0^x g(y) dy \cdots (\ast)$ from the supposition.
From fundamental theorem of calculus, $f$ is differentiable at $x$ and $f'(x)=g(x)$.
But I wonder whether I used the theorem correctly or not.
Fundamental theorem of calculus is here.
Fundamental Theorem of Calculus
If $g$ is continuous on the interval $I\subset \mathbb R$, then $f(x):=\int_a^x g(y) dy$ is differentiable and $f'(x)=g(x)$ for all $x,a\in I.$
In $(\ast)$, $g$ is continuous due to the uniform convergence. But in this case, what plays the role of $I$ ? If I let $I=[-1, x+1]$, then $0,x\in I$, so does $I=[-1,x+1]$ work ?
Postscript
The similar problem is solved here Uniform convergence of derivatives, Tao 14.2.7.. This solved for $f_n : [a,b]\to \mathbb R.$
Now, I wonder whether this holds for $f_n: \mathbb R \to \mathbb R$.
