Convergence that preserves smoothness One of the advantages of uniform convergence is that it preserves continuity (among other properties). Unfortunately, it does not preserve derivability. Is there a convergence mode preserving it?
 A: Let $a$ and $b$ be real numbers with $a<b$. Suppose that $f_n:[a,b]\to\mathbb R$ is continuously differentiable on $[a,b]$ (considering one-sided derivatives at the endpoints). Let $f:[a,b]\to\mathbb R$ and $g:[a,b]\to\mathbb R$ be any functions. 

Claim: If $f_n\to f$ uniformly and $f_n'\to g$ uniformly on $[a,b]$, then $f$ is continuously differentiable and $f'=g$.

Proof: By uniform convergence, $f$ and $g$ are continuous functions. By the fundamental theorem of calculus, one has $$f_n(x)-f_n(a)=\int_a^xf_n'(t)\,\mathrm dt\quad\text{for every $x\in[a,b]$ and every $n\in\mathbb N$.}$$
Next, observe that
\begin{align*}
&\left|f(x)-f(a)-\int_a^xg(t)\,\mathrm dt\right|=\left|f(x)-f_n(x)+f_n(a)-f(a)+f_n(x)-f_n(a)-\int_a^x g(t)\,\mathrm dt\right|\\
=&\,\left|f(x)-f_n(x)+f_n(a)-f(a)+\int_a^x[f_n'(t)-g(t)]\,\mathrm dt\right|\\
\leq&\,|f(x)-f_n(x)|+|f_n(a)-f(a)|+\int_a^x|f_n'(t)-g(t)|\,\mathrm dt\\
\leq&\,\sup_{y\in[a,b]}|f(y)-f_n(y)|+\sup_{y\in[a,b]}|f_n(y)-f(y)|+(x-a)\times\sup_{y\in[a,b]}|f_n'(y)-g(y)|\to 0
\end{align*}
because of uniform convergence. It follows that $$f(x)=f(a)+\int_a^xg(t)\,\mathrm dt.$$ By the fundamental theorem of calculus again, one has that $f$ is differentiable (given that $g$ is continuous) and $f'=g$. $\blacksquare$
