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Suppose that $f:\mathbb{R}\to \mathbb{R}$ satisfying that $f(x+y)+f(x-y)=2f(x)f(y)$ forall $x,y\in \mathbb{R}$ and there exists $x_0$ such that $f(x_0)=-1$. Prove that $f$ is a periodic function.

Here is what I did.

First, letting $x=y=0$, I have $f(0)=0$ or $f(0)=1$. If $f(0)=0$ then $f\equiv 0$, contradiction. Hence $f(0)=1$. Letting $y=0$, I have $f$ is an even function.

After that, letting $y=x_0$ implies that $f(x+x_0)+f(x)=-\big(f(x)+f(x-x_0)\big)$. It means that, if $g(x)=f(x)+f(x-x_0)$ then $g(x+2x_0)=g(x)$ and thus $g$ is a period function.

I have tried to calculate some special values. I have proved that $f(2^nx_0)=1,\forall n\in \mathbb{N}$

I'm stuck here. Can somebody help me?

Mohsen Shahriari
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Dat Tran
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1 Answers1

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Assume $f$ isn't the zero function. Set $y=0$ then $2f(x)=2f(x)f(0)$, whence $f(0)=1$.

Put $x=y=x_0/2$, then $$0=f(x_0)+f(0)=2f(x_0/2)^2,$$ whence $f(x_0/2)=0$. Now, for any $z\in\mathbb{R}$, put $x=z+x_0/2$ and $y=x_0/2$, then $$f(z)+f(z-x_0)=2f(z-x_0)f(x_0/2)=0,$$ so $f(z-x_0)=-f(z)$. Hence, $f(z-2x_0)=f(z)$. We are done.

Mastrem
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