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Find all functions $f:\mathbb{R} \to \mathbb{R}$, that are continuous at the point $x=0$ and satisfy:

$$f (x+y)=f (x)+f (y)+xy (x+y) \ \ \forall x,y \in \mathbb{R} $$

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Well, f(x) = (x^3)/3 satisfies the functional equation and is continuous at 0. –  Adrián Barquero Nov 12 '10 at 20:22
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btw, Mirzodaler: I am guessing you are just posting some interesting challenge problems you have come across. Please try to provide the source of the problems, whenever you can. –  Aryabhata Nov 12 '10 at 20:38
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1 Answer 1

Let $\displaystyle g(x) = f(x) - x^3/3$

Then we have $\displaystyle g(x+y) = g(x) + g(y)$ which by continuity at $\displaystyle 0$ implies $\displaystyle g(x) = kx$.

This gives $\displaystyle f(x) = x^3/3 + kx$ which satisfies the original equation for any $k$.


For a proof that $\displaystyle g(x+y) = g(x) + g(y)$ and $\displaystyle g$ continuous at $\displaystyle 0$ implies $\displaystyle g(x) = kx$

First notice that $\displaystyle g(0) = 0$ and that continuity at $\displaystyle 0$ implies continuity everywhere.

Then by induction, we can prove that for any integer $\displaystyle n$, $\displaystyle g(n) = ng(1)$.

Which can then be extended to the rationals: $\displaystyle g(r) = rg(1) \ \ \forall r \in \mathbb{Q}$.

Given any real $\displaystyle x$, pick a sequence of rationals $\displaystyle \{r_n\}$ converging to $\displaystyle x$ and use the continuity of $\displaystyle g$ at $\displaystyle x$ and that $\displaystyle g(r_n) = r_n g(1)$.

This is a classic and you should find plenty of literature on it.

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I guess after seeing Adrian's comment your work became easy –  anonymous Nov 12 '10 at 20:28
    
@Chandru: I didn't see that comment until you pointed out! –  Aryabhata Nov 12 '10 at 20:29
    
Sorry! Your answer got posted minutes after that was said so i thought you might have got the idea. Anyway +1 nice solution. –  anonymous Nov 12 '10 at 20:30
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