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Note: I'm not asking for a solution or a hint relating to the problem itself, I'm just describing it.

Assume $f: \Bbb R^n \rightarrow \Bbb R^m$ is a continuous function. Furthermore, the equation $$f(x + y) = f(x) + f(y)$$ holds for every $x, y \in \Bbb R^n$. Show that $f$ is a linear function.

So, a friend of mine proved this the following way, and although she did not use the continuity (which makes the proof wrong anyway), I wondered what's so wrong about proving linearity like this in general.

Assume $s \in \Bbb R$. Then $$f(sx) = f\bigg(\overbrace{x + x + \dotsb + x}^{s\text{ terms}}\bigg) = \overbrace{f(x) + f(x) + \dotsb + f(x)}^{s\text{ terms}} = s f(x)$$

I told her that this couldn't be true since she assumes that $|s| \ge 1$, it wouldn't work out for $- 1 < s < 1$ (or would it?). But is this the only reason? And if it is indeed: Wouldn't that mean that we could indeed prove scalar multiplication (and thus: linearity) if we already showed the additivity of a function and could assume that $|s| \ge 1$?

Mike Pierce
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Julian
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    More importantly: writing $sx$ and the sum of $s$ terms, all equal to $x$, only works when $s \in \mathbb{N}$! For linearity, you need $f(sx) = sf(x)$ for all $s$ in the field specified (presumably the real numbers). You argument doesn't show, for example, that $f(\sqrt{2}x) = \sqrt{2} f(x)$. And this is where continuity comes in handy! – StackTD Jun 16 '16 at 16:33
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    Useful: Continuous and additive implies linear – StackTD Jun 16 '16 at 16:36
  • Ha, that's interesting! You're right, this would only work for natural numbers, but since the natural numbers aren't a vector space anyway, we couldn't be confronted with cases like this. – Julian Jun 16 '16 at 16:41
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    I'm not sure what you mean with N not being a vector space. Your argument is not bad, it's just not "finished". You can extend N to Q with similar reasoning, but not to R (without continuity). I converted my comments into an answer. – StackTD Jun 16 '16 at 17:05
  • Like I said in the note at the beginning: This question isn't about the task itself, I simply added it pro forma. The question was about why linearity can't be shown in ways like these, and linearity requires vector spaces, and $\Bbb N$ is not a vector space. Thus: One is not able to prove linearity like this in general. That's what you showed me. :-) Thanks for that! – Julian Jun 16 '16 at 17:10

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For the scalar multiplication, you need $f(sx) = sf(x)$ for all $s \in \Bbb R$. Your argument uses:

Assume $s \in \Bbb R$. Then $$f(sx) = f\bigg(\overbrace{x + x + \dotsb + x}^{s\text{ terms}}\bigg) = \overbrace{f(x) + f(x) + \dotsb + f(x)}^{s\text{ terms}} = s f(x)$$

But writing $sx$ as the sum of $s$ terms, all equal to $x$, only works for $s \in \mathbb{N}$. You argument doesn't show (yet) why $f(\tfrac{1}{2}x) = \tfrac{1}{2}f(x)$. You can however extend your argument to $s \in \mathbb{Q}$ without using the fact that $f$ is continuous, but then you still don't have, for example, $f(\sqrt{2}x) = \sqrt{2} f(x)$. To then close the 'gap' from rationals to reals, continuity of $f$ comes into the picture.

Related: Continuous and additive implies linear.

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