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When a function is additive,

$$\color{blue}{f(x+y)=f(x)+f(y)},$$

you can extend the property to the product by an integer

$$f(nx)=f(x+x+\cdots x)=f(x)+f(x)+\cdots f(x)=nf(x),$$

then to the product by a rational number

$$f(\frac pqx)=\frac qqf(\frac pqx)=\frac1qf(px)=\frac pqf(x),$$

and, presumably by exploiting continuity, to the product by a real number.

So why are all proofs of linearity decomposed in a proof of additivity and a proof of commutativity with scalar multiplication

$$\color{blue}{f(\lambda x)=\lambda f(x)}\ ?$$

Are there conditions on $f$ such that the second part can be omitted ?

Update:

Regarding continuity, one may reason as follows:

By additivity $$\|f(x)-f(x_0)\|=\|f(x-x_0)\|$$ can be made arbitrarily small, as

$$\|f(r(x-x_0))\|=|r|\|f(x-x_0)\|,$$ where $r$ is a rational.

Hence, any additive function $f(x)$ is continuous.

Is that correct ? What are required hypothesis ?

  • I think that you can use only one step with: $f(\lambda_{1}x_{1} +\lambda_{2}) = f( \lambda_{1}x_{1}) + f( \lambda_{2}x_{2}). Hence, if you prove that equality,it should prove linearity. But i am not sure,while we used it only in vectors subsets and sets for now. – MathIsTheWayOfLife Jun 27 '15 at 10:17
  • An additive function need not be continuous, at least if we assume the axiom of choice (pick any non-constant $\mathbb{Q}$-linear map from $\mathbb{R}$ to itself). – Tobias Kildetoft Jun 27 '15 at 10:51
  • @TobiasKildetoft: what is the flaw in the $\delta,\epsilon$ argument then ? (Sorry, your comment is a little too high-level for me.) –  Jun 27 '15 at 11:04
  • It might be clearer if you actually wrote it up with the $\delta$'s and $\varepsilon$'s. – Tobias Kildetoft Jun 27 '15 at 11:11
  • What you are doing is essentially the following: You fix some $x$ and then conclude $f(rx)=r f(x)\to 0$ as $r\to 0$. But this holds (if we do not know that $f$ is continuous) only for rational multiples of $x$, you get no information on irrational multiples of $x$, no matter how close to the origin. – PhoemueX Jun 27 '15 at 12:01
  • @PhoemueX: understood, thanks. –  Jun 27 '15 at 12:20
  • How exactly do you get from $\mathbb Z$-linearity to $\mathbb Q$-linearity? For example, it is clear that $\frac12f(v)$ and $f(\frac12v)$ both be solutions to $x+x=f(v)$, but who says this equation has only one solution? – hmakholm left over Monica Jun 27 '15 at 16:25
  • @HenningMakholm: by $\mathbb Z$-linearity, $2f(v/2)=f(2(v/2))=f(v)$, hence $f(v/2)=f(v)/2$. –  Jun 27 '15 at 16:40
  • @YvesDaoust: ah, OK. – hmakholm left over Monica Jun 27 '15 at 17:45

2 Answers2

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As you noted yourself, there are two conditions for which the second part can not be omitted:

  1. If you are considering a different field than $\Bbb{F} = \Bbb{R}$. (For example $\Bbb{F} = \Bbb{C}$), since then in general $\Bbb{Q}$ will not be dense in $\Bbb{F}$.

  2. If you do not require $f$ to be continuous.

Conversely, if your base field is $\Bbb{F}=\Bbb{R}$ and if you know/assume that $f$ is continuous, then showing additivity suffices.

PhoemueX
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  • You can conclude that if the source space, i.e. the space on which $f$ is defined is finite dimensional. Otherwise, you cannot. For example, consider the space of all sequences $(x_n)_n$ with $x_n=0$ for all but finitely many $n$, equipped with the sup-norm. Let $f$ be the unique linear functional mapping $\delta_n$ to $n$. Then $f$ is not continuous. – PhoemueX Jun 27 '15 at 10:36
  • Sorry for the confusion, I have added an update to explain my thinking. –  Jun 27 '15 at 10:48
  • @PhoemueX Would you have an example of non-continuous function $f$ such that for all $x\in\mathbb{R}^n$, $f(\lambda x)=\lambda f(x)$? – anderstood Sep 18 '17 at 20:06
  • @anderstood: I am not sure I understand what you are asking: If you want $f(\lambda x) = \lambda f(x)$ for rational $\lambda$, note that $\Bbb{R}$ is infinite dimensional as a $\Bbb{Q}$ vector space. Choose an infinite $\Bbb{Q}$-linearly independent set $(x_n)_n$ with $|x_n|=1$, and extend it to a $\Bbb{Q}$-basis $(x_i)_i$ of $\Bbb{R}$, and define a $\Bbb{Q}$-linear map $f : \Bbb{R}\to\Bbb{Q}$ by requiring $f(x_n)=n$ for $n\in \Bbb{N}$, and $f(x_i)=0$ for $i \in I \setminus \Bbb{Q}$. It is not hard to see that this is discontinuous (one reason: the image is not connected). – PhoemueX Sep 18 '17 at 20:23
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If $f$ is continuous, the commutative property can be omitted and vice versa.

As your proof, $f(x+y)=f(x)+f(y)$ can deduce to $f(\lambda x)=\lambda f(x)$ for $\lambda\in \mathbb{Q}$. If you want to extent the conclusion to $\mathbb{R}$, the $f$ must be continuous by using a rational sequence to approximate any real number.

Lion
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  • If the space on which $f$ is defined is finite dimensional, this is true. Otherwise, $f$ need not be continuous. – PhoemueX Jun 27 '15 at 10:37