# Rate of convergence of mollifiers // Sobolev norms

Following up to the question raised here, 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.

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 $$c(\varphi)=\frac12\int z^2\varphi(z) dz.$$
• Well ... sometimes you just imagine things harder than they are. Your proof works the same way to get that exactness on $\mathbb{R}_k[X]$ leads to an estimate of order $k+1$. I'll look for the converse later. Thanks. Jul 9 '19 at 20:45
I stick to the case $$d=1$$ for simplicity and assume that $$\varphi$$ is supported in $$[-1,1]$$. Let $$N$$ be the maximal integer for which $$f\mapsto f\star \varphi$$ is exact on $$\mathbf{R}_N[X]$$. Exactness of the previous map on $$\mathbf{R}_N[X]$$ is equivalent (looking at the value at $$0$$) to \begin{align*} \forall k\in\{0,\cdots,N\},\quad \int_{\mathbf{R}} \varphi(t)t^k \mathrm{d} t =0. \end{align*} The proof given in Bazin's answer adapts to show that for $$f\in\mathscr{D}(\mathbf{R})$$ \begin{align*} \|f-f\star \varphi_\delta\|_p \lesssim \delta^{N+1} \sup_{|\alpha|=N+1} \|\partial^\alpha f\|_p. \end{align*}
Claim : $$\|f-f\star \varphi_\delta\|_p \lesssim \delta^\alpha N(f)$$ is not possible for any $$\alpha> N+1$$ and (semi-)norm $$N$$ (asymptotically as $$\delta\rightarrow 0$$).
Indeed, by the same Taylor expansion we have (using the exactness on $$\mathbb{R}_N[X]$$) for any $$f\in\mathscr{D}(\mathbf{R})$$ $$\begin{multline*} (f\star \varphi_\delta)(x) - f(x) = \int_{\mathbf{R}} \frac{f^{(N+1)}(x)}{(N+1)!}(\delta z)^{N+1}\varphi(z)\,\mathrm{d}z\\+\int_{\mathbf{R}} \int_0^1 \frac{(\delta z)^{N+2}}{(N+1)!} (1-\theta)^{N+1} f^{(N+2)}(x+\theta \delta z)\varphi(z)\,\mathrm{d}\theta\,\mathrm{d}z. \end{multline*}$$ In particular, if $$\eta\in\mathscr{D}(\mathbf{R})$$ equals to $$1$$ on $$[-2,2]$$ and $$f(x):=x^{N+1}\eta(x)$$, then for $$|h|<1$$, we have for $$x\in[-1,1]$$ \begin{align*} f\star\varphi_\delta(x)-f(x) = \delta^{N+1} c(\varphi), \end{align*} from which we infer $$\|f\star\varphi_\delta -f\|_p \geq \delta^{N+1} c(\varphi)2^{1/p}$$.