Theorem. Consider a sequence of differentiable functions {f_n} that converge pointwise to f, whose derivatives Df_n converge uniformly to Df. Then f is differentiable with derivative Df.
Proof.
The "big idea" is that the consequent of the differentiability criterion for f can be bounded by the consequent of the differentiability criterion for f_n (which we know is true), plus the difference between f_n and f (which is arbitrarily small)
Fix ε.
∀n ∃δ_n s.t. ||z − x|| < δ_n → ||f_n(z) − f_n(x) − Df_n(x)[z − x]|| ≤ ε||z − x||
Let δ := inf{δ_n}.
∀x ∃N₁ s.t. n ≥ N₁ → ||f_n(x) − f(x)|| < ε
∃N₂ ∀x ∀y s.t. n ≥ N₂ → ||Df_n(x)[y] − Df(x)[y]|| < ε.
(note quantifier order, uniform convergence means that ∀x is on the inside)
Let N := max{N₁, N₂}.
Suppose ||z − x|| < δ, and n > N.
Add fancy zero to ||f(z) − f(x) − Df(x)[z − x]||
to get ||f(z) + (f_n(z) − f_n(z)) − f(x) + (f_n(x) − f_n(x)) − Df(x)[z − x] + (Df_n(x)[z − x] − Df_n(x)[z − x])||
which commutes and associates to
||(f(z) − f_n(z)) + (− f(x) + f_n(x)) + (− Df(x)[z − x] + Df_n(x)[z − x]) + (f_n(z) − f_n(x) − Df_n(x)[z − x])||
which is triangularly
≤ ||f(z) − f_n(z)|| + ||f_n(x) − f(x)|| + ||Df_n(x)[z − x] − Df(x)[z − x]|| + ||f_n(z) − f_n(x) − Df_n(x)[z − x]||
< ε + ε + ε + ε||z − x||
?!?!?
This isn't the right form; to show differentiability, we need to end on "< ε||z − x||", but the bounds on the differences between sub-n and the limit don't have a ||z - x||
We definitely can't use ||z - x|| in the bounds on the f_n - f terms because pointwise convergence isn't strong enough, but I also don't think we can use it in the Df_n(x)[y] − Df(x)[y] term, either
