2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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. $

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $\|f\star\varphi_\delta -f\|_p \geq \delta^{N+1} c(\varphi)2^{1/p}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.