$\textbf{Proposition:}$ Let $U\subseteq \mathbb{R}^n$ be open and $f:U\to\mathbb{R}^k$ be a totally-differentiable function. Then the derivative map $$df:U\to \mathbb{R}^{k\times n}:x\mapsto df(x)$$ is continuous under the operator norm $||A|| = \sup_{|v|\leq1}\{||Av||\}$.
$\textbf{Proof 1 (incorrect)}$: Since $f$ is totally differentiable, it is a $C^1$-function. Meaning all its partial derivatives exist and are continuous. Take a convergent sequence $x_n\to a$ in $U$, we will show $df(x_n) \to df(a)$ under the operator norm.
Choose $\varepsilon>0$. Since for all $i$ and $j$, the partial derivatives $\partial_jf_i$ are continuous, there are $N_{ij}>0$ such that if $n\geq N_{ij}$ then $||\partial_jf_i(x_n)-\partial_jf_i(a)||<\varepsilon/\sqrt{kn}$. Take $N = \max_{i,j}\{N_{ij}\}$, if now $n\geq N$, then
\begin{align}
||df(x_n)-df(a)||&\leq ||df(x_n)-df(a)||_2 =||df(x_n-a)||_2\\
&=\left(\sum_{i\leq k}\sum_{j\leq n}\partial_jf_i(x_n-a)^2\right)^{1/2}\\
&\leq \left(\sum_{i\leq k}\sum_{j\leq n}\left(\dfrac{\varepsilon}{\sqrt{kn}}\right)^2\right)^{1/2} = \sqrt{\dfrac{kn\varepsilon^2}{kn}} = \varepsilon.
\end{align}
Since $||\cdot||\leq ||\cdot||_2$. Hence, $df(x_n)\to df(a)$.
$\textbf{Question}$: Is this proof correct?
$\textbf{Edit}$: I have rewritten the proof:
$\textbf{Proof 2 (correct)}$: Since $f$ is totally differentiable, it is a $C^1$-function. Meaning all its partial derivatives exist and are continuous. Take a convergent sequence $x_n\to a$ in $U$, we will show $df(x_n) \to df(a)$ under the operator norm.
Choose $\varepsilon>0$. Since for all $i$ and $j$, the partial derivatives $\partial_jf_i$ are continuous, there are $N_{ij}>0$ such that if $n\geq N_{ij}$ then $|\partial_jf_i(x_n)-\partial_jf_i(a)|<\varepsilon/\sqrt{kn}$. Take $N = \max_{i,j}\{N_{ij}\}$. If now $n\geq N$, then
\begin{align}
||df(x_n)-df(a)||&\leq ||df(x_n)-df(a)||_2\\
&=\left(\sum_{i\leq k}\sum_{j\leq n}\big(\partial_jf_i(x_n)-\partial_jf_i(a)\big)^2\right)^{1/2}\\
&\leq \left(\sum_{i\leq k}\sum_{j\leq n}\left(\dfrac{\varepsilon}{\sqrt{kn}}\right)^2\right)^{1/2} = \sqrt{\dfrac{kn\varepsilon^2}{kn}} = \varepsilon.
\end{align}
Since $||\cdot||\leq ||\cdot||_2$. Hence, $df(x_n)\to df(a)$.
Best Answer
Proof 2 is correct. Proof 1 is not, since $df$ needn't be linear.
The whole statement can be reduced to something much more elegant. Consider the following without proof:
Using this fact, it is easily seen (by picking some appropriate norm) that for the normed space of matrices, if the entries of the matrix are continuous, so too is the matrix.
Hence, since $f$ is $C^1$, the linear map represented by the matrix $(\partial_jf_i)_{ij}$ will be continuous since its entries are continuous.
Furthermore, the following claim is then a corollary:
$\textbf{Proof}$: one implication is proved above. Likewise, suppose $df$ is continuous. By picking some appropriate norm (say, the $||\cdot||_2$-norm), one can see that if the matrix is continuous, so too are the entries. Hence, for all $i$ and $j$, the functions $\partial_jf_i$ are continuous. (Existence of the partial derivatives is assumed)