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I'd like to find a function such that $f(x+y) = f(x)^2f(y)^2.$ Yes I realize initial conditions should be specified, but I don't have that luxury, I'm just looking for any function at all and then set the constant in the family of solutions later.

Here's my approach:

First, differentiate by $x$ to obtain

$f'(x+y) = 2f(x)f'(x)f(y)^2.$

Next I set $x = 0$ to find

$f'(y) = 2f(0)f'(0)f(y)^2.$

This is now an ODE which I solve for $f(y) = \frac{2f(0)f'(0)}{c_1 - x}.$

Now the problem is that $f(y+x) = \frac{2f(0)f'(0)}{c_1 - y - x} \neq \frac{2f(0)f'(0)}{c_1 - y} \cdot \frac{2f(0)f'(0)}{c_1-x}.$

Where did this technique go wrong?

daw
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StackQuest
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  • You solved the ODE incorrect. The solution of $f' = cf$ is $f(x) = f(0)\exp(cx)$. – martini Jan 29 '22 at 20:24
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    I made a typo by forgetting the square and corrected it. – StackQuest Jan 29 '22 at 20:25
  • Thx, didn't see this, sry. But: Differentation of $f(y) = c/(c_1 -y)$ gives $f'(y) = c/(c_1-y)^2 = f^2(y)/c$, not $f^2(y) \cdot c$. It should be (if I calculated correctly $f(y) = 1/(c_1 - cy)$. – martini Jan 29 '22 at 20:29
  • I'm not sure I understand what you wrote. $\frac{d}{dx} c f(x) = c f'(x).$ I don't understand why you're suggesting the constant $c$ suddenly reciprocates. In either case, it still doesn't address what's going on with $x$ and $y$. – StackQuest Jan 29 '22 at 20:33
  • Alright only 2 days until I can start a bounty. – StackQuest Jan 29 '22 at 21:21
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    Not every function is differentiable. DO NOT USE DERIVATIVES. – Kavi Rama Murthy Jan 29 '22 at 23:20

1 Answers1

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Where did this technique go wrong?

It doesn't work because you can't assume differentiability, and also can't assume $f(y) \ne 0$.

Instead, if $f(y)=0$ for some $y$ then $f(x)=f^2(y)f^2(x-y)=0$ i.e. the trivial solution $f \equiv 0$.

Otherwise, it means $f(x) \ne 0$ for all $x$, then $f(0+0)=f^2(0)f^2(0) \implies f(0) = 1$, and it follows that $f(x+0)=f^2(x)f^2(0) \implies f(x) = 1$ i.e. the other trivial solution $f \equiv 1$.

dxiv
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  • Thank you for answering. Is there a way to tell "when" differentiating is valid? I would have to suppose $f(y) = 0$ additionally for this method to work? For instance, why is it valid here https://math.stackexchange.com/questions/1289985/polynomial-functional-equation-fxfy2-fxfyfxy?rq=1 ? – StackQuest Jan 30 '22 at 22:16
  • @StackQuest It is only valid to differentiate when you know or otherwise prove that the function is in fact differentiable. The linked question is about polynomial functions, which you know are differentiable. – dxiv Jan 30 '22 at 22:23
  • But how do you know the solution is a polynomial before differentiating and going through the whole process? – StackQuest Jan 30 '22 at 23:13
  • @StackQuest Assuming you ask about the question linked in the previous comment, you know it because the problem states so: "Let f(x) be a ... *polynomial function*". – dxiv Jan 30 '22 at 23:16
  • But what is the basis for making that assumption of the functional equation before being given outside information? If you're given "just" the functional equation without being told it's polynomial, couldn't you deduce it's a polynomial using the differential method? Or is differentiating a completely useless, impractical component to functional analysis? – StackQuest Jan 30 '22 at 23:17
  • @StackQuest What do you mean? That problem is specifically asking you to find a polynomial function. You are making no additional assumption. If the problem did not specify that the solution must be a polynomial function, then it would be a very different problem, and you could not solve it by differentiating. – dxiv Jan 30 '22 at 23:18
  • If you're given that functional equation without being told specifically that the solution is polynomial, could you then deduce that the non-constant solution is a polynomial, without being told explicitly so? Unless of course, there are multiple non-constant solutions in which case, could you deduce that one of finitely many forms is polynomial? – StackQuest Jan 30 '22 at 23:19
  • @StackQuest I don't know the answer to that. But the point is that you could not assume differentiability to solve it. – dxiv Jan 30 '22 at 23:25