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We are given $f(0)=0$. Then when $x+y=0$: $$0=f(-y)+f(-x)+xy$$ Can I now use $x=0$ and obtain: $$0=f(-y)?$$ Is this correct? Is there a better way to solve this equation?

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    $\begingroup$ If you have both $x+y=0$ and $x=0$, then unfortunately $y=0$ so you only learned that $f(0)=0$. $\endgroup$ – vadim123 Dec 26 '15 at 21:54
  • $\begingroup$ OK, so what's the proper method? $\endgroup$ – VanDerWarden Dec 26 '15 at 21:54
  • $\begingroup$ Maybe note that $f(t)=t^2/2$ is a solution. $\endgroup$ – André Nicolas Dec 26 '15 at 21:56
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    $\begingroup$ Taking $y=-x$ we learn that $f(x)+f(-x)=x^2$, for all $x$. $\endgroup$ – vadim123 Dec 26 '15 at 22:01
  • $\begingroup$ And what do we know from that...? $\endgroup$ – VanDerWarden Dec 26 '15 at 22:01
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Set $f(x) = g(x) +\dfrac{x^2}{2}$ then plugging in gives $$\fbox{1}\,g(x+y)=g(x)+g(y).$$ This is Cauchy's functional equation. And under certain regularity assumptions for $g$ you get that $g(x)=ax$. But if none are given then $g$ is just a linear function over $\mathbb{Q}$ the rational numbers and so any of those would be an appropriate solution.

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if you sub $y = -x,$ you get $$0 = f(0) = f(x-x)= f(x) + f(-x) - x^2 \tag 1$$ suppose further assume that $f = ax^2 + bx.$ subbing in $(1),$ gives you $f = \frac12 x^2 + bx$ for any $b.$

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Assuming $f$ differentiable, differentiate the functional equation implicitly wrt. $x$ and $y$ to obtain two simpler equations. From these you obtain $f'(t)=t+c$ after some simple algebraic manipulations.

Assuming $f$ twice differentiable, differentiating twice wrt. $x$ gives $f''(x+y)=f''(x)$, so $f''$ does not depend on $y$. Analogously show that $f''$ does not depend on $x$. So $f''$ is constant. So...

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  • $\begingroup$ So $f = \frac12 x^2 + bx$? $\endgroup$ – VanDerWarden Dec 26 '15 at 22:05
  • $\begingroup$ And how do you differentate function without it's equation? Sorry if this is a stupid question, but I always had to differentiate functions like $f(x)=ax^2+bx$ and so on. $\endgroup$ – VanDerWarden Dec 26 '15 at 22:06
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    $\begingroup$ Just differentiate both sides while treating the other variable like a constant. E.g. differentiating by $x$ gives $f'(x+y)=f'(x)+y$. This implies the above form for $f'$ (why?) and hence $f = \frac12 x^2 + bx+c$, where $c=0$ which you get if plug this in the original equation. $\endgroup$ – Damian Reding Dec 26 '15 at 22:12
  • $\begingroup$ So we treat $f(y)$ as a constant too? $\endgroup$ – VanDerWarden Dec 26 '15 at 22:36
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    $\begingroup$ When differentiating wrt. x - yes. (And when differentiating wrt. y you treat $f(x)$ as a constant) $\endgroup$ – Damian Reding Dec 26 '15 at 22:38
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From the functional equation and the initial condition $f(0)=0$ we get $$ f(x + y) - f(x) = f(0 + y) - f(0) + xy $$ Assuming $f$ is differentiable, dividing by $y$ and taking the limit $y\to 0$ we get $$ f'(x) = f'(0) + x $$ Integration gives $$ f(x) = f'(0)\, x + \frac{1}{2}x^2 $$

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I show that the only solution is $f = \frac{1}{2}x^2 + bx$ (assuming $f$ continuous).

My strategy was to transform $f(x+y) = f(x) + g(y) + xy$ into a linear equation.

Write $f(x+y) = f(x) + g(y) + xy$ as $f(x+y) - \frac{1}{2}(x+y)^2 = (f(x) - \frac{1}{2}x^2) + (g(y) - \frac{1}{2}y^2)$. Substitute $g(z) = f(z) - \frac{1}{2}z^2$. Then we have:

$g(x+y) = x+y$, which is linear, so $g(x) = bx$ (assuming continuity).

Thus $f(z) - \frac{1}{2}z^2 = bz$, so $f(z) = \frac{1}{2}z^2 + bz$.

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  • $\begingroup$ @zxy They posted after I began typing I guess. $\endgroup$ – vhspdfg Dec 26 '15 at 22:26
  • $\begingroup$ Yeah, at the time I began typing the only other answer was like that. I'll edit it to omit that part. Thanks. $\endgroup$ – vhspdfg Dec 26 '15 at 22:33
  • $\begingroup$ yah, the answers here were faster than usual. I'll remove the comments. $\endgroup$ – zyx Dec 26 '15 at 23:12
  • $\begingroup$ That is assuming f is continuous. If not, there are many solutions. $\endgroup$ – DanielWainfleet Dec 27 '15 at 9:03
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$F(x)=2f(x) - x^2$ satisfies $F(x+y)=F(x)+F(y)$ whose solutions are $F(x) = ax$ under the most minimal regularity assumptions (continuous is more than enough).

As several almost identical answers are saying, after subtracting a particular solution of the equation, the inhomogeneous equation $f(x+y)=f(x)+f(y) + (\cdots)$ becomes the well understood homogeneous equation $f(x+y)=f(x)+f(y)$ so that the answer is a linear function added to the particular solution.

To remain agnostic about Axiom of Choice issues one can say that $f(x) = \frac{x^2}{2} + $ (solution of $h(x+y)=h(x)+h(y)$) instead of declaring whether nonlinear solutions to the Cauchy equation exist or not.

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