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For example, given that a linear function is defined by its satisfying the properties $f(x+y)=f(x)+f(y)$ and $f(ax)=af(x)$, would it be okay to only check $f(ax+by)=af(x)+bf(y)$, or maybe even $f\left(\frac ab x+y\right)=\frac abf(x)+f(y)$ and put $c=\frac ab$?

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    $\begingroup$ Sure, why don't you try proving that $f(cx+y)=cf(x)+f(y)$ (for all $c \in F$, $x,y \in V$) suffices to show that $f$ is linear as a function on a vector space $V$ over field $F$, and then you will have answered your own question. $\endgroup$ Feb 20 at 21:46
  • $\begingroup$ Yes, but be careful. Good mathematicians have made mistakes in trying to "combine axioms." E.g., in the first edition of Kelley's General Topology, in defining the concept of a ring, he combined the left and right distributive laws into the single law $(a+b)(c+d)=ac+ad+bc+bd$ which doesn't really work. $\endgroup$
    – user14111
    Feb 20 at 23:27
  • $\begingroup$ @user14111 Interesting, what goes wrong there? $\endgroup$ Feb 22 at 17:59
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    $\begingroup$ @JairTaylor See the answers to this math.overflow question: mathoverflow.net/questions/135527/what-is-a-kelley-ring $\endgroup$
    – bof
    Feb 22 at 22:10
  • $\begingroup$ @bof Very interesting and amusing anecdote, thank you! $\endgroup$ Feb 22 at 23:39

2 Answers 2

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If you have some axioms $A$ and an expression $E$, there are four possibilities:

  1. $A$ and $E$ are logically equivalent, i.e. $A \iff E$, and so anything that satisfies one is guaranteed to satisfy the other.

  2. $E$ is a sufficient condition for $A$, i.e. $E \implies A$, so any system where $E$ is true will also satisfy $A$, but if $E$ is not true you can't make any claim about $A$.

  3. $E$ is a necessary condition for $A$, i.e. $A \implies E$, so if $E$ is not satisfied then $A$ can't be true but not every system that satisfies $E$ necessarily satisfies $A$.

  4. $E$ and $A$ are not directly related, in which case you're out of luck.

So the question is whether $f(ax + by) = af(x) + bf(y)$ is equivalent to $f(x + y) = f(x) + f(y)$ and $f(ax) = af(x)$ (for all vectors $x, y$ and scalars $a, b$). Proving the equivalence in one direction ($A \implies E$) is pretty simple:

$$f(ax + by) = f(ax) + f(by) = af(x) + bf(y)$$

So the axioms definitely imply the given expression. Proving it in the other direction is not too hard but has a couple more steps. For example, if we assume the expression is true and set $a = b = 1$ then we very quickly get

$$f(x + y) = f(1x + 1y) = 1f(x) + 1f(y) = f(x) + f(y)$$

and so you just need to show that $f(ax + by) = af(x) + bf(y) \implies f(ax) = af(x)$.

Of course this assumes that you've already proven the required vector space properties, like $1x = x$ and $0x = 0$.

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I will try to give a more general answer than yes or no. If we are in doubt about such questions, then we can look at if they imply each other. In this case, they are equivalent. For the concrete example: $$"\Rightarrow":$$ Let a function satisfy $(1):f(x+y) = f(x)+(y) $ and $(2):f(\alpha x)= \alpha f(x)$.
Then, $f(\alpha x+y) = f(\alpha x) + f(y) = \alpha f(x) +f(y)$, Where we first used (1) and then (2) So this direction is correct. Now for the other one (which is a bit more tricky): $$"\Leftarrow":$$ Let a function satisfy $f(\alpha x + y) = \alpha f(x) + f(y)$. Then, setting $\alpha=1$, we have $f(\alpha x + y) = \alpha f(x) +f(y) = f(x) + f(y)$, our first property. For the second, we will first show that $f(0) = 0$. This is true because $f(0) = f(0+0)=f(0)+f(0)$, so $f(0)$ must be zero as $0$ is the only real number satisfying $x=x+x$. So, setting $y$ to zero we have: $f(\alpha x + 0) = \alpha f(x) + f(0) = \alpha f(x)$.


We have shown now that, since these two definition imply each other (that is, if one is true, then so is the other, and vice versa), they are equivalent. I hope that you do not just take away a simple yes or no answer, but how to approach these types of questions in general!

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