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Let $f:\mathbb{R}\to\mathbb{R^*}$ be a function such that $f(x+y)+f(x-y)=2f(x)f(y),\forall x,y\in\mathbb{R}$. Prove that $f(x)=1,\forall x\in\mathbb{R}$.

I have managed to prove the following:

1) $f(0)=1$

Set $x=y=0$ so: $f(0)=f^2(0)\Rightarrow f(0)=0\lor f(0)=1$. We cannot have $f(0)=0$, since $f:\mathbb{R}\to\mathbb{R^*}$, so $f(0)=1$.

2) $f(x)=f(-x),\forall x\in\mathbb{R}$

Set $x=0$ so: $f(y)+f(-y)=2f(y)\Rightarrow f(y)=f(-y)$

3) $f(x)=2f^2\left( \frac{x}{2}\right) -1$

Set $y=x$ so: $f(2x)+1=2f^2(x)\Rightarrow f(2x)=2f^2(x)-1$ and setting where x, $x/2$ we get: $f(x)=2f^2\left( \frac{x}{2}\right) -1\Rightarrow f(x)>-1$

I cannot move any further... Any hint?

  • If f is a contant then its derivative should vanish no? But like @IvanNeretin said, is that the only f that satisfies the relation? – Ismail Bello Sep 20 '15 at 11:48
  • This is D'Alembert functional equation. You may search and see that the only solutions are $f(x)=1$ and $f(x)=\cosh(kx)$. – Mohsen Shahriari Sep 27 '15 at 18:27

3 Answers3

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Well, $f(x)=\cosh(a\cdot x)$ for any constant $a$ seems to match the equation, so you may have hard time proving that $f(x)\equiv1$.

As to whether or not this solution (or rather, a family thereof) is unique, I expect it to be so if we require continuity, but that's another story.

Ivan Neretin
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This is known as D'Alembert's functional equation when it is form $\mathbb R$ to $\mathbb R$ and it is known that the only continuous functions $f$ satisfying it are

$$f(x)=0, f(x)=1, f(x)=\cos(kx), f(x)=\cosh(kx)$$

for a certain $k \in \mathbb R$.

Of course, only $f(x)=1, f(x)=\cosh(kx)$ statistify your condition $f(x)>0$ for all $x$.

wythagoras
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Fixed $s\in \mathbb{R}$ and define $g(x):=f(x+s)-f(x-s)$ and consider $$m(x)=f(x)+ag(x)$$ for all $x\in \mathbb{R}$, where $a$ is constant. Then you can show that $$m(x+y)=m(x)m(y)$$ and $$f(x)=\frac{1}{2}(m(x)+m(-x))$$ and going on easy way.

Hint. The following functional equation is refered to as The D'Alembert Functional Equation $$f(x+y)+f(x-y)=2f(x)f(y)..$$ See here for more.

Deliasaghi
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