convergence theorem in differentiation

Question

‎‎‎We know that: Let $‎\lbrace‎ f‎_{n} ‎\rbrace‎‎$‎ be a sequence of differentiable functions defined on $ [0, 1], $‎ and assume $‎\lbrace ‎f'‎_{n}‎\rbrace‎‎ $‎ ‎ ‎converges uniformly to a function ‎$ g $ on ‎$ [0, 1] $. If there exists a point ‎$ x‎_{0}‎\in ‎[0,1]‎ $ ‎such that ‎$ ‎\lbrace ‎f‎_{n} ‎(x‎_{0})‎\rbrace‎‎‎$‎ is a convergent sequence, then $‎\lbrace‎ f‎_{n} ‎\rbrace‎$ ‎‎converges uniformly to ‎$ f$ and and ‎$f'=g . $‎ But if $ f‎_{n}‎\rightarrow ‎f‎ $‎ ‎converges uniformly, this ‎does ‎not ‎imply ‎that‎ $f $ ‎is differentiable and $ f'‎_{n}‎\rightarrow ‎f' ‎$ converges uniformly‎. Is it possible to put some conditions on functions with which conditions can be deduced that $ f‎_{n}‎\rightarrow ‎f‎, $‎ then $ f'‎_{n}‎\rightarrow ‎f' . ‎$

Answer 1

In real analysis, the theorem you quoted is about the best one can do. Convergence (uniform or not) of differentiable functions on an interval $[a,b]$ alone says nothing about the behavior of the derivatives. In complex analysis though, one has much nicer theorems:

If $\{f_n\}$ is a sequence of holomorphic functions on an open set $U\subset\Bbb{C}$ such that the sequence converges uniformly on compact subsets of $U$ to a function $f$, then $f$ is holomorphic on $U$ and $f_n'\to f$ uniformly on compact subsets of $U$.

Another way of saying this is as follows. Let $\mathcal{C}(U)$ be the space of continuous complex-valued functions on $U$, and $\mathcal{H}(U)$ the space of holomorphic functions on $U$. Equip $\mathcal{C}(U)$ with the topology of uniform convergence on compact subsets, and $\mathcal{H}(U)$ the subspace topology. The above result can then be stated as

The space $\mathcal{H}(U)$ is a closed vector subspace of $\mathcal{C}(U)$, and the differential operator $\frac{d}{dz}$ maps $\mathcal{H}(U)$ linearly and continuously into itself.

This goes to show how nicely behaved holomorphic functions are. This theorem follows from the fact that a function $g:U\to\Bbb{C}$ is holomorphic on $U$ if and only if $g$ is continuous and for each $z\in U$, and sufficiently small $r>0$ we have \begin{align} g(z)=\frac{1}{2\pi i}\int_{|\zeta-z|=r}\frac{g(\zeta)}{\zeta-z}\,d\zeta, \end{align} in which case $g'(z)=\frac{1}{2\pi i}\int_{|\zeta-z|=r}\frac{g(\zeta)}{(\zeta-z)^2}\,d\zeta$.

In complex analysis, this (Cauchy's) integral formula provides the connection between functions and its derivatives. Said differently, we're using the fact that in general integrals behave (relatively) nicely with respect to limiting operations to conclude that for the complex case, derivatives also have nice properties.

In real analysis, the differential operator is an unbounded linear map between many of the function spaces we're interested in, so this also provides a more abstract perspective for why we shouldn't expect convergence of functions alone (plus some minor conditions) to imply any sort of relationship on their derivatives.