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According to the question mentioned here, it seems that there is no function $f(x)$ such that the functional equation

$$f(x+y)=f(x)+f(y)-(xy-1)^2$$

can hold. Motivated by this question, I found it interesting to somehow extend the question.

What conditions are required for a given function $g(x)$ such that there exists a function $f(x)$ that can satisfy the following equality $$f(x+y)=f(x)+f(y)+g(xy)$$ where $f(x)$ and $g(x)$ are real valued functions of real variable.

Any hint or help is appreciated. :)

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Hosein Rahnama
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    Trivial observation: for each fixed $a \in \mathbb{R}$, the functions $f(x)=ax^2$ and $g(x) = 2ax$ satisfy the required constraint. I guess the question, then, is really: do other examples exist, and if so, are they continuous? – goblin GONE Jan 28 '16 at 18:17
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    In fact, $ax^2+b$ and $2ax-b$ seems to work. – goblin GONE Jan 28 '16 at 18:18

2 Answers2

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If $f$ is not differentiable, we still have $f((x+y)+z)=f(x+(y+z))$, from which follows

$$g(xy)+g(xz+yz)=g(xy+xz)+g(yz)\tag{1}$$

By change of variables

$$x=\sqrt{\frac{ab}{c}}, \quad y=\sqrt{\frac{ca}{b}}, \quad z=\sqrt{\frac{bc}{a}}\tag{2}$$

this becomes

$$g(a)+g(b+c)=g(a+b)+g(c)\tag{3}$$

whenever $abc>0$.

By change of variables $(a,b,c)\to(a+b,-b,b+c)$ we can also move $-b$ across so it holds for all $abc\neq0$. Then

$$g(a+b)+g(b-b)=g(a)+g(b)\tag{4}$$

so $h(x)=g(x)-g(0)$ satisfies $$h(a+b)=h(a)+h(b)\tag{5}$$ This has the well-known solutions $h(x)=kx$, and some very discontinuous ones.

Hosein Rahnama
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Empy2
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    The relation between $(x,y,z)$ and $(a,b,c)$ is not obvious. Maybe it could be useful if you added some explanation. – Mohsen Shahriari Jan 28 '16 at 20:23
  • Thanks for the nice answer. I added some more details to your solution. But I still have some questions about it. Firstly, where the condition $abc \gt 0$ comes from? I think each of $a,b,c$ should be positive $a,b,c>0$. :) – Hosein Rahnama Jan 28 '16 at 21:08
  • One more thing, according to your solution, if $g(x)$ is continuous then it must be a linear function of $x$ right? – Hosein Rahnama Jan 28 '16 at 21:14
  • Yes, $g(x)$ is linear for rational values of $x$, so the irrational values follow by interpolation. – Empy2 Jan 29 '16 at 03:45
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If $f$ is differentiable then taking the derivative with respect to $x$ we get

$$f'(x+y)=f'(x)+yg'(xy).$$ Now, taking the derivative with respect to $y$ one has $$f''(x+y)=g'(xy)+xyg''(xy).$$ If $y=-x$ then $$g'(-x^2)-x^2g''(-x^2)=f''(0).$$ If we denote $t=-x^2$ we have $$tg''(t)+g'(t)-f''(0)=0.$$ Solving the ODE we get $$g(t)=f''(0)t+a\ln t+b,$$ where $a,b$ are arbitrary constants. If we want that $g$ is defined for all $t$ it must be $a=0.$ So $$g(t)=f''(0)t+b.$$ Now, from $$f''(x+y)=g'(xy)+xyg''(xy)$$ we get that $$f''(x+y)=f''(0).$$ So the second derivative of $f$ must be constant. That is, $f$ is a second degree polynomial. It only remain which polynomials of degree two can be a solution. It is a straightforward computation to check that any $f(x)=ax^2+bx+c$ is a solution with $g(x)=2ax-c.$

mfl
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  • It only remain which polynomials of degree two can be a solution. So? Which ones? – Najib Idrissi Jan 28 '16 at 18:33
  • @NajibIdrissi I have edited the answer. – mfl Jan 28 '16 at 18:35
  • @mfl I know this is a silly question but how do you differentiate $g (xy) $ wrt $x $ I used $u=xy $ and the chain rule but that gives me $\frac {dg}{du}y $ not $g'(xy)y $ I'd like to follow the rest of your answer. Thanks. – Karl Jan 28 '16 at 19:42
  • @Karl The derivative of $g(xy)$ with respect to $x$ is $\frac{dg}{du}\cdot \frac{du}{dx}.$ Now, $\frac{dg}{du}=g'$ and $\frac{du}{dx}=y.$ – mfl Jan 28 '16 at 19:45
  • @mfl I am reading $g'$ as $\frac {dg}{dx} $ why is it allowed to be $\frac {dg}{du} $ if $f'=\frac{df}{dx} $ thanks for your help. – Karl Jan 28 '16 at 19:53
  • @Karl If you write $u(x)=xy$ and you uses the chain rule $\frac{d(g\circ u)}{dx}=\frac{dg}{du}\frac{du}{dx}$ then $\frac{dg}{du}=g'(u).$ Note that the derivative of $g(u)$ with respect to $u$ is $g'(u).$ – mfl Jan 28 '16 at 19:56
  • @mfl Thanks I need to think this through. I'm having a mental block at the moment. I can't get round idea that the prime notation appears to be used to first denote the derivative wrt $x$ and then in another term wrt $u$ I can see $u$ is a function of $x$ Sorry for asking too many questions. – Karl Jan 28 '16 at 20:12
  • Thanks for the contribution. How did you obtain the third equation? Can you elaborate please? :) – Hosein Rahnama Jan 28 '16 at 20:51
  • I had a typo. It should say $f''(0).$ Anyway, if $y=-x$ then $f''(x+y)=f''(0)$ $g'(xy)=g'(-x^2)$ and $xyg''(xy)=-x^2g''(-x^2).$ Thus $f''(0)=g'(-x^2)-x^2g''(-x^2).$ – mfl Jan 28 '16 at 22:52
  • @mfl: What linear combination of the first and the second equations do you consider to get the third one? I think there is still $f'(0)$ missing! :) – Hosein Rahnama Jan 29 '16 at 07:35
  • The third equation is just the second one when $y=-x.$ – mfl Jan 29 '16 at 08:28
  • @mfl: (+1), I got it now. :) – Hosein Rahnama Jan 29 '16 at 16:31
  • @mfl I posted a follow up question as I'm still having difficulties aligning the notations. Thanks for your help – Karl Apr 01 '16 at 18:58