Let $\varrho \in L^\infty_{loc}$, where the set of discontinuity points is a Lebesgue null set, and $\varrho * \varphi$ is a polynomial with maximum degree of $m$ $\forall \varphi \in C_c^\infty$. Now we know that $$0=\frac{\mathrm{d}^{m+1}}{\mathrm{d} x^{m+1} } (\varrho * \varphi)(x) = \int_{\mathbb{R}} \varrho(x-y) \frac{\mathrm{d}^{m+1}}{\mathrm{d}x^{m+1}}\varphi^{(m+1)}(y) dy.$$
I need this for an elaboration of a proof. There they state that with the help of Distribution theory I can follow that $\varrho$ is a polynomial. Has someone maybe an idea how this works and can give me some input? Thank you very much
Let $\varphi$ be a mollifier and let $\varphi_n(x) = 2^{n} \varphi(2^n x)$ so that $\rho \ast \varphi_n \to \rho$ almost everywhere and as a distribution.
Since $\rho\ast \varphi_n$ is a polynomial of degree at most $m$, we can write $$\rho \ast \varphi_n(x) = \sum_{k \le m} a_{k,n} x^k.$$
Now let $\psi \in C_c^\infty$ be a function such that $$\int \psi(x) x^k dx = \begin{cases} 1 \text{ if } k = 0 \\ 0 \text{ if } 0 < k \le m. \end{cases}$$
Then $\langle \rho \ast \varphi_n , D^k \psi \rangle_{L^2} = (-1)^k a_{k,n}$ for each $k \le m$ so that $$(-1)^k a_{k,n} \to \langle \rho, D^k\psi \rangle_{L^2} =: (-1)^k a_k$$ by the convergence as a distribution. Therefore, for almost all $x$, $$\rho(x) = \lim_{n \to \infty} \sum_{k \le m} a_{k,n} x^k = \sum_{k \le m} a_k x^k.$$