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So, I'd like to prove that the function $f(xy) = f(x) + f(y)$ is strictly increasing within the domain

$f: [0, \infty] \rightarrow\mathbb R $

and that

$f(x) > 0 $ when $x > 1$

and

$f(x) < 0 $ when $x < 1$

Now, I understand that by using the definition of strictly increasing function, it is possible to observe the derivative and see if derivative function has an extreme point. However, I just cannot understand how it could be done in this case.

I'm very thankful for all help in advance!

Thomas Andrews
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Maskurate
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    Note that if your function is actually defined at $0$ then it must be the constant $0$, as $f(x\times 0)=f(0)=f(x)+f(0)$ for all $x$. Moreover it isn't clear what you mean by "the function". If all you are assuming is the functional equation, then the function is not uniquely defined. – lulu Oct 11 '22 at 11:56
  • Certainly $f(1)$ must be zero, because $f(1) = f(1\cdot 1) = f(1) + f(1) = 2f(1).$ Additionally, $f(1/x) = -f(x)$, because $f(x) + f(1/x) = f(x\cdot\frac 1 x) = f(1) = 0$. – CiaPan Oct 11 '22 at 12:03
  • I am not sure if one can prove those statements without additional assumptions on $f$, e.g. continuity. – Martin R Oct 11 '22 at 12:07
  • Related: https://math.stackexchange.com/q/3860102/42969 – Martin R Oct 11 '22 at 12:09
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    Can you edit your post for clarity? As you can see from the comments, the assumptions you have in mind are not clear. – lulu Oct 11 '22 at 12:11
  • Voting to close the question as it is not clear what you are asking. – lulu Oct 11 '22 at 12:26

1 Answers1

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In view of the comment by lulu I will assume that the domain is $(0,\infty)$.

Also, if $f$ satisfies the given equation so does $-f$. So I assume that $f(x) > 0 $ when $x > 1$

and

$f(x) < 0 $ when $x < 1$ is part of the hypothesis, not the conclusion.

If $y>x\geq 1 $ then $y=zx$ where $z=\frac y x >1$. Hence, $f(y)=f(zx)=f(z)+f(x) >f(x)$. So $f$ is strictly increasing in $[1,\infty)$. I will let you use a similar arguemt for $(0,1]$. Since $f$ is negative in $(0,1)$ and positive in $(1,\infty)$ (and $f(1)=0$) it follows that it is strictly increasing in $(0,\infty)$.

geetha290krm
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  • Do you assume that $f(z) > 0$ for $z > 1$ or do you prove that? – Martin R Oct 11 '22 at 12:07
  • @MartinR Please read the comments in the beginning of my answer now. – geetha290krm Oct 11 '22 at 12:14
  • As I understand the question, the goal is to show that $f$ is increasing with $f(x) > 0$ for $x> 1$ and $f(x) < 0$ for $x < 1$. So “$f(x) < 0$ when $x < 1$” is not part of the hypothesis. – Martin R Oct 11 '22 at 12:17
  • Also the given equation can be transformed into the Cauchy function equation (see the link in my comment at the question) and that can have quite irregular solutions. – Martin R Oct 11 '22 at 12:19
  • You cannot prove that. If $f$ satisfies those conditions then $-f$ will not. OP has not stated the question correctly I am making an obvious, necessary modification. @MartinR – geetha290krm Oct 11 '22 at 12:20
  • @MartinR With my interpretation of the problem you won't have irregular solutions. – geetha290krm Oct 11 '22 at 12:21
  • Thank you for the explanation! I was (still am) a bit confused on this particular question. It would seem straightforward but it is actually quite mind-boggling. Thank you again! – Maskurate Oct 11 '22 at 13:47
  • @MartinR Not everyone who posts a question has to have a good knowledge of English. Now that OP has approved my answer it is clear that OP could not state the problem clearly but what was meant by OP is exactly what I have written. In view of this, this is not a duplicate either and I resent the attitude of all those who have donvote the question. – geetha290krm Oct 11 '22 at 23:20