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Determine all functions $f \colon\mathbb{R}\to\mathbb{R}$ satisfying the following two conditions:

(a) $f(x +y) + f(x - y) = 2 f(x) f(y)$ for all $x, y\in\mathbb{R}$;

(b) $\lim\limits_{x\to\infty}f(x) = 0$.

I found this problem in IMO 1985 longlist. I have been able to figure out that there is one solution of the form $f(x) = 0$ for all $x$.

Any other function satisfying the above has to have $f(0) = 1$ and $f(x) = f(-x)$.

However I don't know how to proceed.

Any thoughts?

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Why is it that $f(0)=1$ if $f\neq0$? –  miniBill Jun 14 '12 at 9:26
    
@miniBill: If $f\neq 0$ choose $x$ with $f(x)\neq 0$. Then take $y=0$ and apply the cancellation law to equation (a). –  Shane O Rourke Jun 14 '12 at 11:25
    
Understood, thanks :) –  miniBill Jun 14 '12 at 15:00

3 Answers 3

up vote 12 down vote accepted

For any $x_0,y\in\mathbb{R}$, let $x=y+x_0$, then we have

$f(2y+x_0)+f(x_0)=2f(y+x_0)f(y)$

Hence,

$f(x_0) = 2f(y+x_0)f(y)-f(2y+x_0),\quad \forall y\in\mathbb{R}$

Consequently,

$\displaystyle f(x_0) = \lim_{y\to\infty}\left[f(x_0)\right] = \lim_{y\to\infty}\left[2f(y+x_0)f(y)-f(2y+x_0)\right] = 0$.

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This argument generalizes much more easily than mine. –  Erick Wong Jun 14 '12 at 5:24
    
I don't understand why you can say that $f(x_0)=\lim_{y\to\infty}\left[...\right]$ –  miniBill Jun 14 '12 at 9:15
    
@miniBill The second to last line is true for all $y$ (in $\mathbb R$), so you can take the limit of $y$ to infinity and get the same value, but we know that value is also $0$. –  Mark Hurd Jun 14 '12 at 10:19
    
I'm still not convinced that $f(x) = g(x,y) \forall y$ implies $f(x) = \lim_{y\to\infty}g(x,y)$ –  miniBill Jun 14 '12 at 10:44
1  
@miniBill You just take the limit of both sides. You don't need anything for that - in the end both sides are equal. Then on the LHS the sequence is constant and on the RHS we may use the assumption. –  Simon Markett Jun 14 '12 at 10:47

Using $x=y$, we have $f(2x) = 2f(x)^2 - 1$. Pick $X>0$ large enough so that $|f(x)| < 1/2$ for all $x \ge X$. Then $|f(2X)| > 1/2$, a contradiction.

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Your argument is a nice one! –  user17762 Jun 14 '12 at 5:30
    
With $y=x$, the condition is $f(2x)+f(0)=2f(x)^2$, not what you wrote. Taking $x\to\infty$ in the correct condition, we find that $f(0)=0$. –  Jonathan Sep 5 '12 at 23:21
    
@Jonathan Did you not notice that $f(0)=1$? –  Erick Wong Sep 6 '12 at 1:10
    
That's not a condition of the problem. You can show that $f(0)=f(0)^2$ so either $f(0)=0$ or $f(0)=1$. The limit condition rules out the possibility of $f(0)=1$. –  Jonathan Sep 6 '12 at 1:43
1  
@Jonathan Now you're just being deliberately obtuse. You can just as easily show that $f(x) = f(x) f(0)$ so either $f$ is identically zero or $f(0)=1$. Notice the OP already deduced this. –  Erick Wong Sep 6 '12 at 4:27

This is really a comment, not a solution, but I don't have that option. If one only considers the functional equation (a), then there are lots of interesting solutions, namely $\cos ax$ and $\cosh ax$. If one "cheats" and uses imaginary arguments, one can combine these in one formula. By the standard method of inserting a power series and comparing coefficients, one can show that these are the only analytic solutions. Presumably, minimal smoothness conditions on a solution would imply analyticity. However, there are pathological non-measurable solutions. Of course, these solutions are not relevant for the original question since they do not vanish at infinity.

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