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Let $f:\mathbb{R} \to \mathbb{R}$ be a function which satisfies $e^xf(y)+e^yf(x)=2e^{x+y}-e^{x-y}$ for all real $x$ and $y$.

If I place $x=y$, I get $f(x)=e^x-\frac{1}{2}e^{-x}$ which does not satisfy the original equation.

Now instead, if I set$x=y=0$, I get $f(0)=\frac{1}{2}$. Now setting $y=0$ in the original equation, I get $f(x)=\frac{e^x}{2}$.

My question is that why do we get incorrect and different solutions if we proceed in the ways mentioned above? Also, what is the correct solution of this functional equation and how can I find it?

Mohsen Shahriari
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Akshit
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    You have correctly shown that no such function exists. – Empy2 Aug 03 '15 at 16:37
  • @Michael: How can I say that no other function exists which satisfies the given equation? – Akshit Aug 03 '15 at 16:43
  • By putting $x=y$, you showed the only possible function is $e^x-\frac12e^{-x}$. No other will work. But nor does that function. – Empy2 Aug 03 '15 at 16:44
  • Also, if I solve other equations like polynomial or trigonometric equations, I don't get solutions which do not satisfy the original equation. But why is this problem occurring here ? If no solution exists, shouldn't I get no solutions at all, instead of incorrect ones? – Akshit Aug 03 '15 at 16:46
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    What other equations? – marty cohen Aug 03 '15 at 16:49
  • If you look for a linear function that goes through $(0,0),(1,1),(2,3)$, you can use just the first two points to show it is $y=x$. But that doesn't work either, so there is no solution. – Empy2 Aug 03 '15 at 16:50
  • @Akshit The main reason why polynomials work and your function does not, is that your function has a self-reference in its definition. These are always dangerous and and lead to absurdities. Hope this answers one of your questions posted in the comments – Shailesh Aug 08 '15 at 22:51

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Suppose $g(x)f(y)+g(y)f(x)=2g(x+y)-g(x-y) $ where $g(0) \ne 0$.

Setting $x=0$, $g(0)f(y)+g(y)f(0)=2g(y)-g(-y) $.

Setting $y=0$, $g(x)f(0)+g(0)f(x)=2g(x)-g(x) =g(x) $ or $g(0)f(x) =g(x)(1-f(0)) $.

Setting $x=y=0$, $g(0)f(0) =g(0)(1-f(0)) $. Since $g(0) \ne 0$, $f(0) = \frac12$ and $f(x) =\frac{g(x)}{2g(0)} $.

Putting this back,

$\begin{array}\\ 2g(x+y)-g(x-y) &=g(x)f(y)+g(y)f(x)\\ &=g(x)\frac{g(y)}{2g(0)}+g(y)\frac{g(x)}{2g(0)}\\ &=\frac{g(x)g(y)}{g(0)}\\ or\\ g(0)(2g(x+y)-g(x-y)) &=g(x)g(y)\\ \end{array} $.

Setting $x=y$, $g(0)(2g(2x)-g(0)) = 2g^2(x) $. Setting $x=0$, $g(0) = 2g^2(0) $. Since $g(0) \ne 0$, $g(0) = \frac12 $.

Therefore $2g(x+y)-g(x-y) =2g(x)g(y) $.

This does not hold for $g(x) = ae^x+b$, so the original problem has no solution.

I don't know how to solve this, so I'll propose it as a question.

And here it is: Are there any solutions to $2g(x+y)-g(x-y) =2g(x)g(y)$ with $g(0) \ne 0$?

marty cohen
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