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Find all the functions $f\colon\mathbb{Q} \to \mathbb{Q}$ such that $f(x+y) + f(x-y) = 2f(x) + 2f(y)$, for all rationals $x,y$.

zyx
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priya
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2 Answers2

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$f(0)=0$ by choosing $x=y=0$. Also by choosing $x=y$ we can see $f(2x)=4f(x)$. Moreover choose $x=0$ and use $f(0)=0$, then you have $f(y)=f(-y)$.

Now by strong induction one can prove that $f(nx)=n^2f(x)$.

Then choose $x$ as $\frac{x}{n}$ in preceding equation, then you get: $$ f(x)={n^2}f \left(\frac{x}{n}\right)\rightarrow\frac{1}{n^2}f(x)=f \left(\frac{x}{n}\right). $$ And finally we have: $$ f \left(\frac{mx}{n}\right)=\frac{1}{n^2}f(mx)=\frac{m^2}{n^2}f(x). $$ Now assume that $f(1)=a$ for some $a\in\mathbb{R}$, then you get $$ f(x)=ax^2 $$ for all rational $x=\frac{m}{n}$.

M. A.
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Arash
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    Indeed $f(\frac{m}{n})=\frac{m^2}{n^2}f(1)$. – Arash Sep 06 '13 at 13:16
  • Well, that was fine. I probably didn't read correctly. So I deleted the comment. However, I don't understand how you say "Moreover choose $y=−x$ and use $f(0)=0$...". Putting that in the equation we get: $ f(0) + f(2x) = 2f(x) + 2f(-x) $. How do you say from this that $f(x) = f(-x)$? – Parth Thakkar Sep 06 '13 at 13:18
  • Ok, understood. Excuse these silly questions! – Parth Thakkar Sep 06 '13 at 13:20
  • You still can prove that the function is even with this choice by using $f(2x)=4f(x)$. Another option is to choose $x=0$ and then the answer is more direct. – Arash Sep 06 '13 at 13:22
  • $f(0) + f(2x) = 2f(x) + 2f(-x)$ with $f(0)=0 $ and $f(2x)=4f(x)$ gives $4f(x)=2f(x)+2f(-x)$ so $2f(x)=2f(-x)$ so $f(x)=f(-x)$ – Julian Wittische Sep 06 '13 at 13:25
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    @Ouistiti, you can use $\text{ \text {the text here} }$ to display textual content within math. And as I said, I understood it already. – Parth Thakkar Sep 06 '13 at 13:27
  • Thanks that will be much useful ! Sorry I did not see that comments changed ! – Julian Wittische Sep 06 '13 at 13:28
  • How can you take $y=x=0$? What if it is a function of the form $f(x)=y$, so what if the point $(0,0)$ does not lie on the function? – Aditya Agarwal Jan 06 '16 at 12:57
  • Your solution is partially right, first prove that the point lies on the function, by taking $y=0$. – Aditya Agarwal Jan 06 '16 at 12:59
  • @AdityaAgarwal, Sorry your comment is not comprehensible. – Arash Jan 06 '16 at 14:21
  • Which one? I am just saying that how do you know that the point $(0,0)$ lies on the function we are looking for? – Aditya Agarwal Jan 06 '16 at 14:32
  • @AdityaAgarwal, I do not KNOW it. It can be simply derived that $f(0)=0$. Note that we know $0$ belongs to the domain. Have you seen the beginning of the answer "$f(0)=0$ By choosing $x=y=0$"? – Arash Jan 06 '16 at 20:43
  • @Arash That is what I am saying. Let there be any function g(x) satisfied by the condition. Is it necessary that the point (0,0) lies on it? Then how can you assume that before hand? – Aditya Agarwal Jan 07 '16 at 01:56
  • @AdityaAgarwal, No it is not! You have to prove it. Again, I have to ask you again: did you understand the beginning of the answer "$f(0)=0$ By choosing $x=y=0$"? – Arash Jan 07 '16 at 11:53
  • @AdityaAgarwal, If you choose $x=0$ and $y=0$ as two "inputs" of the function $f$ in the above equation you get $f(0)+f(0)=2f(0)+2f(0)$ from that we get $f(0)=0$. – Arash Jan 07 '16 at 11:58
  • Okay wait a minute. Suppose there was any random functional equation and its solution was $ax^2+3$, then, if in beginning, you took $x=y=0$, then you wouldn't get all the answers. – Aditya Agarwal Jan 07 '16 at 12:55
  • @AdityaAgarwal, Ok, I lost you. I do not really see how what you say can be translated into an argument related to the current problem. I hope someone else can help. Why $ax^2+3$ pops up here? The whole point of the argument is that for any function $f$ satisfying the condition of original problem $f(0)$ MUST be zero and hence it is impossible to have $ax^2$ as solution! – Arash Jan 07 '16 at 13:29
  • @AdityaAgarwal, I guess the proper way of arguing here is to show what is wrong with my argument about $f(0)=0$. You can either point me to a flaw in the proof or give me a counter example (a function satisfying the functional equation and $f(0)\neq 0$.). – Arash Jan 07 '16 at 13:32
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That equation, which is often called Appolonius' formula or the parallelogram law, is a characterization of homogeneous quadratic forms on a vector space (without solutions of $v+v=0$).

In plane geometry, with $f(\mathbb{x})=|\mathbb{x}|^2$ the squared Pythagorean distance, the formula is the relation proved by Appolonius between the diagonals and sides of a parallelogram. In a vector space, for any dot product [symmetric bilinear form] $B$, the associated 'squared distance' function [the restriction to a quadratic form] $f(x)=B(x,x)$ satisfies the equation, due to

$B(x+y,x+y)=B(x,x) + B(y,y) + 2B(x,y)$.

Conversely, any solution of the equation comes from this construction, and $2B(x,y) = f(x+y) - f(x) - f(y) $ . The posted problem is the case of a 1-dimensional vector space over the field of rational numbers.

zyx
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