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I need some help to find all of the functions which satisfy the equation: $$f(x+y)-f(x-y)=f(x)f(y)$$

Actually I have no idea how to start to solve this problem.

Sebastiano
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MTMH
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    Do you assume $f$ differentiable? Suppose you have such a function. Can you get $f(0)$? A differential equation that $f$ must fulfill? – nejimban Oct 17 '21 at 08:50

1 Answers1

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Clearly $f \equiv 0$ is a solution. We claim that it is the only solution.

Suppose $f \not \equiv 0$, note

$$f(0) - f(0) = f(0)^2 \Rightarrow f(0) = 0.$$

Thus with $y = x$ we have

$$f(2x) = f(x)^2.$$

Thus $f \geq 0$ as $f(x) = f(\frac{x}{2})^2 \geq 0.$

With $y = -x$ we have

$$-f(2x) = f(x)f(-x) \Rightarrow -f(x)^2 = f(x)f(-x).$$

Thus $f(x) \neq 0 \Rightarrow f(x) = -f(-x)$; this means that either $f(x)$ or $f(-x)$ is negative but this contradicts the fact that $f$ is non-negative.

acreativename
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  • Good analysis, but the asker here wants to know all the functions that satisfies the conditions you have mentioned. – abstractnature Oct 17 '21 at 09:14
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    The only solution to $f : \mathbb{R} \rightarrow \mathbb{R}$ that satisfies $\forall x,y\in \mathbb{R}; f(x+y)-f(x-y) = f(x)f(y)$ is $f = 0$. – acreativename Oct 17 '21 at 09:16
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    @acreativename I think you have been advised to have a look at Enforcement of Quality Standards three days ago, in a comment to this answer of yours. Please also note that if you keep answering low quality questions, then your answers will likely be flagged for moderator intervention. This is clearly stated in the above linked meta post. – WhatsUp Oct 17 '21 at 09:22