Let $ f : \mathbb R \to \mathbb R $ be a continuous function that satisfies $$ f ( u + a v ) = f ( u ) + f ( a v ) + P _ n ( u , v ) $$ 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 $.
On
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 $\bf n$
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.