I am wondering if term by term Differentiability still works if we weaken the assumption $\Sigma_{n=1}^{\infty}f_n'(x)$ converges uniformly to a limit $g(x)$ by only requiring convergence, but not uniform convergence. In other words, if there exists a point $x_0 \in [a,b]$ where $\Sigma_{n=1}^{\infty}f_n(x)$ converges, does the series still converge uniformly to a differentiable function $f(x)$ satisfying $f'(x) = g(x)$ on A?
I am almost certain that it won't always work without requiring uniform convergence of $\Sigma_{n=1}^{\infty}f_n'(x)$, I am having trouble coming up with an example to prove the point. Any ideas will be greatly appreciated.
I will show a counterexample with sequences. Let $f_n\colon[0,1]\to\Bbb R$ be $$ f_n(x)=n(n+1)(n+1)\Bigl(\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\Bigr). $$ Then $f_n$ converges pointwise to $0$ on $[0,1)$ and $f_n(1)=n$, so that the convergence is not uniform. The derivatives are $$ f_n'(x)=n(n+1)(n+1)x^n(1-x), $$ and they converge pointwise (but not uniformly) to $0$ on $[0,1]$.