Cauchy's functional equation with polynomial

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function that satisfies $$f(u+av)=f(u) + f(av) + P_n(u,v)$$ where $a$ is a known constant and $P_n(u,v)$ is a polynomial in $u$ and $v$ of degree $n$. Is $f$ a polynomial of degree $n$?

Edit: I'm looking for a solution that doesn't use derivatives but rather clever'' substitutions for $u$ and $v$.

First of all, let asume wlog $a=1$, otherwhise take $\tilde u = u$, $\tilde v = av$ $$f(\tilde u+\tilde v) = f(\tilde u) + f(\tilde v) + P\left(\tilde u, \frac{\tilde v}{a}\right)$$ and $P\left(\tilde u, \frac{\tilde v}{a}\right)$ is also a polynomial of degree $n$ in $\tilde u,\tilde v$. From the functional equation we have $$f(x-y) = f(x) + f(-y) + P(x,-y)$$ Lets take $g\in C_0^\infty(\mathbb R)$ such that $$\int_{\mathbb R} g(y) dy = 1$$,then $$f(x-y)g(y) = f(x)g(y) + f(-y)g(y) + g(y)P(x,-y)$$ Also $f$ is continous so $f\in L_{loc}^1(\mathbb R)$ hence $f*g\in C^\infty(\mathbb R)$ and we can integrate the last equation $$f*g(x) = f(x)+ \int_{\mathbb R}f(-y)g(y)dy + \int_{\mathbb R} g(y)P(x,-y) dy$$ $$f(x)=f*g(x) - \int_{\mathbb R}f(-y)g(y)dy - \int_{\mathbb R} g(y)P(x,-y) dy$$ the three function in the right hand side are $C^{\infty}$ so $f\in C^{\infty}$ and the proof continues as Aditya sugest for example.
Rough Outline: $$f'(u+av)=f'(u)+P'_{n-1}(u,v)$$ Put $v=-u/a$: $$f'(u)=f'(0)-P'_{n}(u,-u/a)$$ $$f(u)=P^{\circ}_{n}(u)+c_1u+c_2\quad \text{where }c_1\text{ and }c_2\text{ are constants}$$ even though $\bf P'_n$ has subscript $\bf n$ it is of degree $\bf n-1$, $\bf P^{\circ}_{n}(u,-u/a)=\int P'n(u)du$ which is of degree $\bf n$
• A similar approach can be applied to the Cauchy functional equation but is it possible to do this without taking derivatives (like some solutions to the Cauchy functional equation where we substitute clever'' choices of $u$ and $v$)?.... I think some of your $v$'s should be replaced by $u$'s. – user103828 Sep 7 '14 at 14:37
• Do you mean $P'_n(u,v)$? – almagest Sep 7 '14 at 14:39
• When you put $v=-u/2$ the $P$ term becomes $P'_n(u,-u/2)$. So how do you integrate it back to $P_n(u,v)$? – almagest Sep 7 '14 at 14:41
• Of course, all you are trying to do is to establish that $f$ is a polynomial of degree $n$, so all that is fixable. – almagest Sep 7 '14 at 14:43
• @almagest tried to fix it$\require{cancel}\cancel{?}$ – RE60K Sep 7 '14 at 14:47