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A real valued function $f(x)$ satisfies the functional equation $f(x – y) = f(x) f(y) – f(a – x) f(a + y)$

Where $a$ is a given constant and $f(0) = 1\;,$ Then prove that $f(2a – x) = -f(x)$.

and calculate all function which satisfy given functional equation.

$\bf{My\; Try::}$ Given functional equation $\displaystyle f(x-y)=f(x)f(y)-f(a-x)f(a+y)..............(1)$

Put $x=y=0$ in equation $(1)$

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

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

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

My Question is How Can I calculate all function which satisfy The Given functional equation.

Help me

Thanks

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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?

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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.

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    $\begingroup$ 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

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