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Let $\mathbb{K}$ represent either $\mathbb{R}$ or $\mathbb{C}$ and let $E$ denote a normed vector space. Then, a map $f:X \subset \mathbb{K} \rightarrow E$ is said to be approximately linear at a point $a \in X$ if if there exists an affine function $g:\mathbb{K} \rightarrow E$ such that $f(a) = g(a)$ and

$$\lim_{x \to a} \frac{\|f(x)-g(x)\|}{|x-a|} = 0$$

My question is, is the presence of the norm and absolute value in the above definition necessary? The expression $x - a$ is just a scalar in $\mathbb{K}$ and $f(x) - g(x)$ is just a linear combination of vectors and so the expression

$$\lim_{x \to a} \frac{f(x)-g(x)}{x-a} $$

is meaningful

In fact, would it not be the case that the normed quotient approaches the number $0$ if and only if the unnormed quotient approaches the vector $0_E$?

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"is the presence of the norm ... necessary?" - well, if $f$ and $g$ are vector-valued, and you're expecting a scalar result (0), well... –  J. M. May 14 '11 at 17:43
    
To be more clear, the normed quotient approaches the number 0 if and only if the unnormed quotient approaches the vector 0. I will update the question accordingly –  ItsNotObvious May 14 '11 at 17:56
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1 Answer 1

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Yes. Since $\|\lambda e\|=|\lambda|\cdot\|e\|$ for any scalar $\lambda$ and any $e\in E$, and if $F\colon \mathbb{K}\to E$ then $\lim\limits_{x\to a}\;F(x) = 0_E\iff \lim\limits_{x\to a} \|F(x)\|=0$ by the definition of limits in a normed space, the two conditions are equivalent.

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