Let $f\in{L_{loc}^{1}}$ such that $\int_{a}^{b}{f(x)}{\phi}^{(n)}(x)dx=0$ for all $\phi\in{C_{0}^{\infty}}(a,b)$. Then how do we show there exists a polynomial $P(x)$ of degree less or equal to $n-1$ such that $f=P$ a.e.?
My try is: I showed the basic case that is if $\int_{a}^{b}{f(x)}{\phi}'(x)dx=0$ for all $\phi\in{C_{0}^{\infty}}(a,b)$ then $f$ is a.e. equal to a constant function. But I have no idea how to generalize this base result to the case for $n$ in general.
Set $X_+=\{x:f(x)>0\}$ and $X_-=\{x:f(x)<0\}$. If any of them, say $X_+$, has positive measure, then it contains inner points and thus an open interval $(c,d)\subset [a,b]$. Now select a non-trivial non-negative $ϕ∈C^∞_0(a,b)$ with support inside $(c,d)$ and draw conclusions from $\int^b_af(x)ϕ^n(x)dx=0$.
Maybe there is something missing in the assumptions.
The corrected assumption is that $\int^b_af(x)ϕ^{(n)}(x)dx=0$ for any $ϕ∈C^∞_0(a,b)$. For the distributional interpretation see the comments.
Try a more elementary approach. Let $b_h$ be the box function that has value $1$ from $0$ to $h>0$. Then the convolution product $b_h*ϕ∈C^∞_0(a,b)$ inherits smoothness from $ϕ$ and the compact support from both factors united. $$ (b_h*ϕ)'(x)=(b_h*ϕ')(x)=\int_0^h b_h(t) ϕ'(x-t)\,dt=\int_{x-h}^x ϕ'(s)\,ds=ϕ(x)-ϕ(x-h) $$ and so $$ \int_{\Bbb R}g(x)\,(b_h*ϕ)'(x)\,dx=\int_{\Bbb R}(g(x)-g(x+h))\,ϕ(x)\,dx=\int_{\Bbb R}((-Δ_h) g)(x)\,ϕ(x)\,dx $$ Repeated application of this identity results in $$ \int_{\Bbb R}f(x)\,((b_h*)^nϕ)^{(n)}(x)\,dx=\int_{\Bbb R}((-Δ_h)^n f)(x)\,ϕ(x)\,dx $$ which means that $Δ_h^n f=0$ a.e. resulting in $f(x+kh)=p_{x,h}(x+kh)$ a.e., where $p_{x,h}$ is a polynomial of degree less than $n$.
Next to prove: $p_{x,h}$ is, a.e., independent of $x$ and $h$. Perhaps include some smoothing kernel in the convolution product, making the smoothed version of $f$ continuous and thus one polynomial everywhere...