Finding all continuous functions such that $f^2(x + y) − f^2(x − y) = 4f(x)f(y)$

Question

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?

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. — Gerry Myerson, Jan 26 '13 at 05:22
$y = 0$ causes x to be free, so the general solution has $y = 0, x \in \mathbb{R}$ — SSumner, Jan 29 '13 at 20:22

score 9 · Accepted Answer · answered Jan 26 '13 at 05:55

9

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?

answered Jan 26 '13 at 05:55

L. F.

8,498

Ross Millikan · Answer 2 · 2013-01-26T05:42:39.547

0

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

edited Jan 26 '13 at 05:42

answered Jan 26 '13 at 05:33

Ross Millikan

374,822

1

Then, when you simplify by taking the square root of both sides, you get $f(2x) = 2f(x)$, which I already have. – SSumner Jan 26 '13 at 17:47

Finding all continuous functions such that $f^2(x + y) − f^2(x − y) = 4f(x)f(y)$

2 Answers2