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Let $\lambda$ be a nonzero real constant. Find all functions $f,g : \mathbb R \rightarrow \mathbb R$ that satisfy the functional equation for all $x,y \in\Bbb R$: $$f(x + y) + g(x-y) = \lambda g(x) f(y)$$

I tried this: $y=0$ implies $f(x)+g(x)$=$\lambda g(x)f(0)$. Here we have two cases depending on $f(0)$:

  1. $f(0)=0$ or
  2. $f(0) \neq 0$.

If this is true, what next?

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    $\begingroup$ Is there more to the last sentence? $\endgroup$ – vadim123 May 14 '15 at 16:55
  • $\begingroup$ Also try $y = x$, $y = -x$, both $x,y = 0$, etc. $\endgroup$ – Pieter21 May 14 '15 at 17:02
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    $\begingroup$ In that it seems as if you are new here, you might consider making the title of your question more specific. This might generate a bit more interest, since everyone posting a question is asking for help. $\endgroup$ – user12802 May 14 '15 at 17:05
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Let $x=y=0$ then the equation becomes $$ f(0)+g(0)=λg(0)f(0) $$ Case I. If $f(0)=0$ then $g(0)=0$ and vice versa. Setting $y=0$ in the original equation gives $$ f(x)+g(x)=\lambda g(x)f(0)=0 \\ \therefore\:f(x)=-g(x) $$ Similarly setting $x=0$ we get $f(y)=-g(-y)$ and so $f(x)=-g(x)=-g(-x)$. Now setting $y=x$ in the equation we get $$ f(2x)+g(0)=\lambda g(x)f(x) \\ \therefore\:f(2x)=-\lambda f^2(x) \\ \therefore\:f(x) \text{ must be of the form } ca^x $$ Substituting in above equation we get that either $c=0 \mbox{ or } c=-1/\lambda$. $c=0 \mbox{ gives } f(x)=g(x)=0$ for all $x$. For the other value of $c$ we would need $a=0$ to satisfy $f(0)=0$. Even if we assume that $f(x)$ is discontinuous at $x=0$ plugging $f(x)=ca^x$ in the above equation results in $a^y=0$ for all $y$. Hence $a$ has to be zero.

Case II. If $\lambda f(0)=1$ this would result in $f(0)=0$. So $g(0)=0$ and the analysis becomes the same as above

Case III. $f(0)\ne0$, $g(0)\ne0$, $\lambda f(0)\ne1$ and $\lambda g(0)\ne1$. From setting $x=0,y=0$ we get $$ \lambda = \frac{1}{f(0)}+\frac{1}{g(0)} $$ Set $y=0$ and the equation becomes $$ f(x)+g(x)=\lambda g(x)f(0) \\ \therefore\:f(x)=g(x)(\lambda f(0)-1) \\ =g(x) \frac{f(0)}{g(0)} $$ Similarly setting $x=0$ yields $$ g(-y)=f(y)(\lambda g(0)-1) \\ \therefore\:g(x)=f(-x)(\lambda g(0)-1) \\ =f(-x)\frac{g(0)}{f(0)} $$ Combining the above two equations yields $$ f(x)=f(-x) $$ The original equation then becomes $$ f(x+y)+f(x-y)\frac{g(0)}{f(0)}=\lambda f(x)\frac{g(0)}{f(0)}f(y) $$ Replacing $y$ by $-y$ we get $$ f(x-y)+f(x+y)\frac{g(0)}{f(0)}=\lambda f(x)\frac{g(0)}{f(0)}f(y) $$ This implies that either $f(0)=g(0)$ or $f(x+y)=f(x-y)$. The second scenario results in $f(x)=g(x)=2/\lambda$. If $f(0)=g(0)$ the equation reduces to $$ f(x-y)+f(x+y)=\lambda f(x)f(y) $$ The only way to satisfy this is if $f(x)=\text{a constant}$. Hence in this case too $f(x)=g(x)=2/\lambda$.

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  • $\begingroup$ How did you conclude $f(x)=ca^x$ from $f(2x)=-\lambda f(x)^2$? $\endgroup$ – Mohsen Shahriari Dec 2 '15 at 12:34
  • $\begingroup$ $f(x)=\frac2\lambda\cos(ax)$ and $f(x)=\frac2\lambda\cosh(ax)$ satisfy $f(x+y)+f(x-y)=\lambda f(x)f(y)$. $\endgroup$ – Mohsen Shahriari Dec 2 '15 at 12:38
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Few suggestions that I haven't worked out completely, but in a sketch looked promising:

Use $x=0$ to find $f(y) = ...$ in terms of $g(y), \lambda, g(0) $. $(1)$

Use $y=0$ to find another $f(x) = ...$. $(2)$

With just renaming the parameters from both equations, you could get a formula for $g(x)$.

Or you could replace $f(x)$ in your equation with the result of $(1)$ or $(2)$.

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