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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?

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    What operator norm? We have $f(x), f(a) \in \Bbb R^m$, so we are talking about some $\Bbb R^m$ norm here (any of which would do for the definition). Anyhow, if $\|\cdot\|_{op}$ is replaced with $\|\cdot\|_m$ (some norm on $\Bbb R^m$), then yes, you get the definition. – 2017-01-07
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    @OpenBall What's wrong with using the operator norm? That's what my lecture notes have. – 2017-01-07
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    The operator norm is a norm defined on the space of bounded linear operators from a certain normed space into another, so I don't see what it might mean here. – 2017-01-07
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    what's your definition of operator norm? – 2017-01-07
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    Does it make a difference if we consider $x \rightarrow D f(x)$ instead? The derivative of $f$. – 2017-01-07
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    @mavavilj yes it does, because then we are considering an element of the space of bounded linear functions from $\Bbb R^n$ to $\Bbb R^m$, for which we need the operator norm. – 2017-01-07
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    @OpenBall How do you know the derivative is bounded? – 2017-01-07
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    @mavavilj by the definition of $Df$. – 2017-01-07
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    @OpenBall Which is? – 2017-01-07
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    Which is.. $f$ is differentiable at $x$ $\iff$ there exists a bounded linear map $Df(x): \Bbb R^n \to \Bbb R^m$, such that $\lim_{h\to 0} \frac{\|f(x+h)-f(x) - Df(x)h\|}{\|h\|} = 0$. – 2017-01-07

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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.