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. $

  • $\begingroup$ 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. $\endgroup$ – Ayman Moussa Jul 9 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 $\|f\star\varphi_\delta -f\|_p \geq \delta^{N+1} c(\varphi)2^{1/p}$.


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