Continuity using operator norm? Is this the "standard" definition for continuity using vector functions?
Let $f: G \rightarrow \mathbb{R}^m$, $G \subset \mathbb{R}^n$.
Then consider $Df$, the derivative of $f$.
$Df$ is continuous at $a$, if
$$\lim_{x \rightarrow a} || Df(x)-Df(a) ||_{op} = 0$$
Why does one need the operator norm here?
 A: For $f:U \subset \Bbb R^n \to \Bbb R^m$ differentiable at $x\in U$, we have, by definition, that $Df(x)$ is a bounded linear map $\Bbb R^n \to \Bbb R^m$, i.e. an element of $L(\Bbb R^n,\Bbb R^m)$, which is equipped with the operator norm. If $f$ is differentiable on $U$, then $Df:U \to L(\Bbb R^n,\Bbb R^m), x\mapsto Df(x)$ defines a function.
In general, if $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ are normed vector spaces, the operator norm on $L(X,Y)$ is defined for $L\in L(X,Y)$ by:
$$\|L\|_{op} = \inf\{A>0: \forall x\in X, \|L x\|_Y \le A\|x\|_X\} = \sup_{x\neq 0_X} \frac{\|Lx\|_Y}{\|x\|_X}$$
Also, in general, if $U$ is an open subset of a normed vector space $(E,\|\cdot\|_E)$, and $(F,\|\cdot\|_F)$ is a normed vector space, and $g: U \to F$ and $a\in U$, then we say that $\lim_{x\to a}g(x) = L \in F$ if:
$$\forall \epsilon >0, \exists \delta >0: \forall x \in U, \|x-a\|_E < \delta \implies \|g(x)-L\|_F < \epsilon$$
We say that $g$ is continuous at $a$ if $\lim_{x\to a}g(x) = g(a)$. This is equivalent to saying that $\lim_{x\to a}\|g(x) - g(a)\|_F = 0$. 
I hope this helps.
