Following up to the question [raised here](https://mathoverflow.net/questions/57410/rate-of-convergence-of-smooth-mollifiers), I am searching for a reference (or a simple argument) to establish (in the whole space) the following (suggested) equivalence : 



> 
> Given $N\_1$ and $N\_2$ two (homogeneous spaces semi-) norms with
>  scaling exponents $t$ ans $s$ (as in the answer given by @fedja in the
>  cited post), a convolution operator $f\mapsto f\star \varphi\_\delta$
>  is exact on polynomials of degree equal to or less than $r=t-s$ if and
>  only if it satisfies $N\_1(f-f\star \varphi\_\delta) \lesssim \delta^r
> > N\_2(f)$.
> 
> 
> 


For instance, if $\varphi$ has her moments of order $\leq 3$ vanishing, do we have $\|f-f\star\varphi\_\delta\|\_2 \lesssim \delta^3 \sup\_{|\alpha|=3} \|\partial^\alpha f\|\_2$ on $H^3(\mathbb{R}^d)$ ? And why, if these moments are not vanishing, such a result is hopeless ?


I am quite sure that this result (maybe a bit modified in its statement) should be doable *via* Fourier analysis, however I would very much appreciate a "direct" proof of such an inequality, in the spirit of the ordre $1$ case which is based on $\|f-\tau\_\delta f\|\_2 \leq \|\nabla f\|\_2$, that can be proven by Taylor formula.



Rate of convergence of mollifiers // Sobolev norms
You have, say with $\varphi\ge 0$ even, with integral 1,
$$
(f\ast \varphi\_\delta)(x) -f(x)=\int \bigl(f(x+\delta z)-f(x)\bigr)\varphi(z) dz.
$$
As a consequence, we get with Taylor's formula with integral remainder,
$$
(f\ast \varphi\_\delta)(x) -f(x)=\int \int\_0^1(1-\theta)f''(x+\theta \delta z)\delta^2 z^2\varphi(z) d\theta dz,
$$
So that, by translation invariance of the $L^2$-norm and Jensen's inequality
$$
\Vert f\ast \varphi\_\delta-f\Vert\_{L^2}\le c(\varphi)\delta^2\Vert f''\Vert\_{L^2},
$$
where 
$
c(\varphi)=\frac12\int z^2\varphi(z) dz.
$


