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This weird question was given by my professor as a part of my assignment :

$\textbf{Question :}$

“Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that it satisfies $f(x + y) = x f(\frac{1}{y}) + y f(\frac{1}{x})$, whenever $x$ and $y$ are both irrational numbers. Then prove that $f(0)$ is always $0$.”

$\textbf{My attempt :}$

Since $0$ is rational, we can’t let any of the variables to be $0$. So I tried the substitution $y = -x$ to get $f(0) = x f(\frac{1}{-x}) - x f(\frac{1}{x})$.

I tried proving f to be an even function after this, so that $f(0)$ becomes $0$. But I couldn’t do so or proceed further anyhow.

Can somebody kindly provide me hints or solutions for this problem ?

Vinay Karthik
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  • What is $.$ here? – Marco Feb 27 '24 at 16:38
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    Hint: Take the equations for $x = y = t$ and $x= y = 1/(2t)$. What can you conclude? – Robert Israel Feb 27 '24 at 16:44
  • . is the simple product of real numbers. I shall edit it once – Vinay Karthik Feb 27 '24 at 16:48
  • "(0) is always 0" is not understandable. – user Feb 27 '24 at 16:55
  • @user It could be better phrased but it is easily understandable: "for every such function $f$, $f(0)=0$". – Anne Bauval Feb 27 '24 at 17:05
  • @RobertIsrael Shouldn't your comment be an answer? – Anne Bauval Feb 27 '24 at 17:11
  • @RobertIsrael, After considering your hints, I got $2t^2 = 1$. Is that what you meant ? So I let $y = -x$ and $x = \frac{1}{√2}$ to solve this. Thank you for your help ! – Vinay Karthik Feb 27 '24 at 18:04
  • @VinayKarthik Can you post that as a full solution, or edit it into your answer? – Calvin Lin Feb 27 '24 at 19:34
  • @AnneBauval I did not see this possibility. This means it was not easily understandable. – user Feb 27 '24 at 19:44
  • @RobertIsrael Is the conclusion that the function is identically $0$ correct? – user Feb 27 '24 at 22:26
  • @VinayKarthik No, what you get is not $2t^2=1$. – Anne Bauval Feb 27 '24 at 22:55
  • @CalvinLin , Here is my solution. Let $x = y = \frac{1}{\sqrt{2}}$, in the original equation. After simple algebra, we get $f(\sqrt{2}) = 0$. Similarly $f(-\sqrt{2}) = 0$. Now let $x = \frac{1}{\sqrt{2}}$ and $y = -x$ to conclude. – Vinay Karthik Feb 28 '24 at 03:35
  • @user, Even I had the same hunch as yours. So should I ask the question of whether f is identically 0 as a new question or edit the current question ? Any suggestions ? I am new member, so I am quite unsure. – Vinay Karthik Feb 28 '24 at 05:15
  • This is a complicated question. If you are sure that you found the correct answer you can publish it. Doing this I would thank @RobertIsrael for the hint. If you are unsure you can edit the question, give the alleged solution and ask if it is correct. – user Feb 28 '24 at 07:23
  • @VinayKarthik Great. Instead of leaving it as a comment, can you post it as an answer (down below), or edit it into your question? – Calvin Lin Feb 28 '24 at 15:10

2 Answers2

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With $x = y = t$ (any irrational) you get $$ f(2t) = 2 t f(1/t) $$ and with $x = y = 1/(2t)$ you get $$f(1/t) = f(2t)/t$$ Substitute the first into the second and it says $$ f(1/t) = 2 f(1/t)$$ from which you conclude $f(1/t) = 0$, i.e. $f$ is $0$ on all irrationals. Then use $y = -x$ to get $f(0) = 0$.

Robert Israel
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Here is my solution, after taking suggestions from @RobertIsrael in the comments.

Let $x = y = \frac{1}{\sqrt{2}}$, in the original equation. After simple algebra, we get $f(\sqrt{2}) = 0$. Similarly we get $f(-\sqrt{2}) = 0$ by taking $x = y = \frac{-1}{\sqrt{2}}$ . Now let $x = \frac{1}{\sqrt{2}}$ and $y = −x$ to conclude.

Vinay Karthik
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  • It would be correct to explain the "simple algebra" here and not in the comment to the previous answer. – user Feb 29 '24 at 11:05