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I've been working on the following homework problem:

Find all continuous functions $f : \mathbb{R} → [0,∞)$ such that $f^2(x + y) − f^2(x − y) = 4f(x)f(y)$ for all real numbers $x, y$.

The first problem I have is I am not sure what is meant by "finding all continuous functions". Can someone show me what the ending statement should look like?

Secondly, I'm not sure what to do with the relations I've derived. The most helpful are:

$f(0) = 0$

$f(x) = f(-x)$

$f(2x) = 2f(x)$

I also know that:

$f(2) = 2f(1)$

$f(3) = \frac{3}{2}f(2)$

$f(4) = 2f(2) = 4f(1)$

This leads me to believe that $\frac{f(a)}{f(b)} = \frac{a}{b}$, but I don't know (a) if this is helpful and (b) how to prove it.

Can someone clarify what exactly the object is, and point me towards the next step?

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    $\begingroup$ One such function is $f(x)=x$. That doesn't help you find the general solution, I just thought someone should point out the one easy example. $\endgroup$ – Gerry Myerson Jan 26 '13 at 5:22
  • $\begingroup$ $x=2$ and $y=0$ work fine for me... $\endgroup$ – Alan Simonin Jan 26 '13 at 5:36
  • $\begingroup$ $y = 0$ causes x to be free, so the general solution has $y = 0, x \in \mathbb{R}$ $\endgroup$ – SSumner Jan 29 '13 at 20:22
  • $\begingroup$ If $0<x,y<\infty$, the solution is $y=x$. $\endgroup$ – Takahiro Waki Apr 30 '17 at 22:32
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The condition $f(x)\geq 0$ seems to kill all non-trivial possibilities: since $f(0)=0$ we let $y=-x:$

$$-f^2(2x)=4f(x)f(-x)$$

However $f(x)\geq 0$, so $f(x)=0$. Are you absolutely sure of the $\mathbb{R}\to [0,\infty)$ bit?

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  • $\begingroup$ @Rustyn, I just looked at the title, and missed the non-negativity condition. $\endgroup$ – Gerry Myerson Jan 26 '13 at 11:50
  • $\begingroup$ I checked, and I have the problem stated exactly as it was presented. $\endgroup$ – SSumner Jan 26 '13 at 17:48
  • $\begingroup$ @SSumner In that case it is most definitely $0$. $\endgroup$ – L. F. Jan 26 '13 at 17:54
  • $\begingroup$ Interesting. Thanks for your help. I will ask my professor if they was an error. If not, I'll accept $\endgroup$ – SSumner Jan 26 '13 at 17:56
  • $\begingroup$ No error, so apparently the only possible solution is indeed $f(x) = 0$. Thanks for your help $\endgroup$ – SSumner Jan 28 '13 at 17:31
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One thing that helps is to find useful values. If you set $y=0$ you find $f(0)=0$. Then if $x=y$ you get $f^2(2x)=4f^2(x)$

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    $\begingroup$ Shouldn't that be $f^2(2x)$ on the LHS? $\endgroup$ – L. F. Jan 26 '13 at 5:36
  • $\begingroup$ Then, when you simplify by taking the square root of both sides, you get $f(2x) = 2f(x)$, which I already have. $\endgroup$ – SSumner Jan 26 '13 at 17:47

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