Consider a sequence of functions $f_n(x)$, which converges to $f(x)$ pointwise on the interval $(0,1)$. The functions $f_n(x)$ and the limit function $f(x)$ are differentiable, with derivatives $f'_n(x)$ and $f'(x)$ respectively.
Assume that the sequence $f'_n(x)$ converges to a function pointwise. Because convergence of the derivatives is not necessarily uniform, the usual theorem for term-by-term differentiability cannot be applied to give $\lim_{n \rightarrow \infty} f'_n(x) = f'(x)$. However,
If it is known that both $f_n(x)$ and $x f'_n(x)$ converge uniformly on $[0,1]$, does that imply that $\lim_{n \rightarrow \infty} f'_n(x) = f'(x)$ on $(0,1)$?
I think the answer is yes. But it strikes me a bit that I get the desired result side-stepping the need for uniform convergence of $f'_n(x)$.
So I'll post my proof and please let me know if there's any mistake in it. Or if the result is well-known, can you point me to a relevant theorem?
Best Answer
Yes, the implication holds
Proof
Since $f_n(x)$ converges to $f(x)$ uniformly on the compact interval $[0,1]$, $x f_n(x)$ converges to $x f(x)$ uniformly on that interval.
The derivative of $xf_n(x)$ is $f_n(x) + x f'_n(x)$, which by the assumptions converges uniformly on $[0,1]$. Therefore, the referred theorem can be applied to the sequence of functions $x f_n(x)$. As a result, the derivatives of these functions converge to the derivative of $x f(x)$; that is, $f_n(x) + x f'_n(x)$ converges to $f(x) + x f'(x)$. This implies that $x f'_n(x)$ converges to $x f'(x)$. For $x \in (0,1)$, dividing by $x$ shows that $f'_n(x)$ converges to $f'(x)$.