A real valued function $f$ satisfies the functional equation $$f(x - y) = f(x) f(y) - f(a - x) f(a + y) \tag 1 \label 1$$

Where $a$ is a given constant and $f(0) = 1$. Prove that $f(2a - x) = -f(x)$, and find all functions which satisfy the given functional equation.

My Try:

Put $x=y=0$ in equation \eqref{1}.

$\implies f(0)=f(0)^2-f(a)\cdot f(a)\implies f(a)^2=0\implies f(a)=0$.

Now Put $x=a$ and $y=x-a$.

$\implies f(2a-x)=f(a)\cdot f(x-a)-f(0)\cdot f(x) = -f(x)$.

My question is how can I find all function which satisfy the given functional equation.

Help me.



Partial progress:

Let $P(x,y)$ be the property that $f(x-y) = f(x)f(y)-f(a-x)f(a+y)$.

Then, $P(0,0)$ gives $f(0) = f(0)f(0)-f(a)f(a)$ which yields $f(a) = 0$.

Also, $P(a,x)$ gives $f(a-x) = f(a)f(x)-f(0)f(a+x)$ which yields $f(a-x)=-f(a+x)$.

Then, $P(x,x)$ gives $f(0) = f(x)f(x)-f(a-x)f(a+x)$ which yields $f(x)^2 + f(x+a)^2 = 1$ (*).

Using (*), we have $f(x)^2+f(x+a)^2 = 1$ and $f(x+a)^2+f(x+2a)^2 = 1$ for all $x \in \mathbb{R}$.

Hence, $f(x+2a) = \pm f(x)$ for all $x \in \mathbb{R}$. Therefore, $f(x+4a) = f(x)$ for all $x \in \mathbb{R}$.

It's pretty clear that $f(x) = \cos\dfrac{\pi x}{2a}$ is a solution. Now, can we show that there aren't any more?

| cite | improve this answer | |

It seems that if $a \ne 0$, then $f(x) = \cos\frac{\pi }{2a}x$, and if $a = 0$, there is no solution!

Sketch of proof: The case $a=0$ is obvious. So let that $a\ne0$. With a change of variable the equation can be change to $$ f(x+y) = f(x)f(y)-f(\pi/2-x)f(\pi/2-y). $$ Let $g(x)=f(x)+i f(\pi/2 -x)$. Then one can easily see that $g(x+y)=g(x)g(y)$. From this and $g(0)=1 $ one can see that $g(x) = e^x$ and the rest is straightforward.

| cite | improve this answer | |
  • 2
    $\begingroup$ But $f(x) = 0$ does not satisfy $f(0) = 1$. So $a = 0$ has no solutions. $\endgroup$ – JimmyK4542 Jul 1 '14 at 3:36
  • $\begingroup$ @JimmyK4542 Thanks Jimmy! I have corrected it! :) $\endgroup$ – Mohammad Khosravi Jul 1 '14 at 3:39
  • $\begingroup$ @user7530! I have written a sketch of solution. :) $\endgroup$ – Mohammad Khosravi Jul 1 '14 at 4:04
  • $\begingroup$ en.wikipedia.org/wiki/Cauchy%27s_functional_equation $\endgroup$ – Mohammad Khosravi Jul 1 '14 at 4:19
  • $\begingroup$ Do we know that $g$ must be continuous? Otherwise, we could have other solutions. $\endgroup$ – JimmyK4542 Jul 1 '14 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.