uniform continuity, differentials

Question

Let $\{f_n\}_{n=1}^\infty$ be a sequence in $C^1(U)$ where $U \subset \mathbb{R}^d$ is open. Suppose $f_n \to f$ uniformly on compact subsets of $U$. Assume further that $df_n \to A$ in the same sense where $A$ is a function on $U$ taking values in the $d\times d $ matrices. How would I go about proving that $f \in C^1(U)$ and that $df = A$ in $U$? I know that limits of uniformly convergent sequences of continuous functions are continuous. For the derivatives, should I use the definition of the linearization with the error $r(x, h)$?

Answer 1

For the derivatives it is not advisable to use the definition of the linearization with the error $r(x, h)$. Instead, write for any $x, y \in D \subset U$ where $D$ is an open disk in $U$,$$f_n(x) - f_n(y) = \int_0^1 df_n(y + t(x-y))\,dt(x-y).$$Passing to limits we obtain$$f(x) - f(y) = \int_0^1 A(y + t(x-y))\,dt(x-y).$$We know that $A$ is continuous. This means that for $y$ fixed and any vector $v$, the limit$$\lim_{h \to 0} h^{-1}(f(y+hv)- f(y)) = A(y)v$$exists, whence $df(y) = A(y)$ as desired.