Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking for a reference for the $L_p$ error of the difference of a Sobolev function and its convolution with a band-limited mollifier.

The type of estimate that is quoted in a paper without a source is as follows:

Consider $f\in W_p^k(\mathbb{R})$, and $\phi$ is a band-limited function ($\widehat{\phi}$ is supported on the interval $[-\pi/(2h), \pi/(2h)]$ for some small $h>0$). Their estimate was $$\|f-(f\ast\phi)\|_{L_p(\mathbb{R})}\leq Ch^k|f|_{W_p^k} = Ch^k\|f^{(k)}\|_{L_p}$$

The author simply says by the "usual error estimate from mollification by a band-limited mollifier."

Any book or paper that proves a similar estimated would be appreciated.

share|cite|improve this question
up vote 3 down vote accepted

Some references for the rate of $L^p$ approximation in terms of Sobolev norms are given here. Some insightful remarks on the rate of convergence of mollified functions are found here.

Below I give a proof of the estimate (not in your interpretation, which misses important points).

Let $\varphi$ be as in the paper you are reading. For $h>0$ let $\varphi_h(x)=h^{-1}\varphi(x/h)$. That the support of $\widehat{\varphi_h}$ is contained in $\{\xi: |\xi|\le (\pi+\varepsilon)/h\}$ is not of much help. What really matters is that $\widehat {\varphi_h}=1$ on $\{\xi: |\xi|\le c/h\}$ where $c=\pi-\varepsilon$. Since the Fourier transform of $f*\varphi_h$ agrees with $\widehat f$ when $|\xi|\le c/h$, the transform of $f-f*\varphi_h$ vanishes for such $\xi$.

As usual, the case $p=2$ is easier to deal with. Let $M$ be the supremum of $|\widehat \varphi| $ (which can be arranged to be $1$). Then
$$ \begin{split} \|f-f*\varphi_h\|_{L^2}^2 & = \int_{|\xi|\ge c/h} |\widehat f_1(\xi)|^2 d\xi \\ & \le (h/c)^{2k} \int_{\mathbb R} |\xi|^{2k} |\widehat f_1(\xi)|^2 d\xi \\ &\le (h/c)^{2k} M \int_{\mathbb R} |\xi|^{2k} |\widehat f(\xi)|^2 d\xi \\ &= (h/c)^{2k} M |f|_{W^{k,2}}^2 \end{split} \tag1$$

The idea for the general case $1<p<\infty$ is about the same: $f-f*\varphi_h$ has only high frequencies of $f$ (those above $c/h$), which are magnified in $f^{(k)}$ by factors of at least $(c/h)^k$. But to relate the Fourier transform to the $L^p$ norm, we need the Littlewood-Paley decomposition of $f$: $$f=\sum_{j\in \mathbb Z} f_j$$ where $ f_j =f*\varphi_{2^{-j}}-f*\varphi_{2^{-j+1}}$. The important point here is that $\widehat{f_j}$ is supported in a roughly dyadic annulus of size $2^{j}$. These frequencies get magnified by about $2^{jk}$ in the $k$th derivative. The Littlewood-Paley theorem says that $\|f\|_{L^p}$ is comparable to the $L^p$ norm of the square function of $f$: $$\|f\|_{L^p}\approx \left\|\left(\sum_{j\in\mathbb Z} |f_j|^2\right)^{1/2}\right\|_{L^p} \tag{LP}$$ Use (LP) for $f^{(k)}$ and for $f-f*\varphi_h$: $$ \|f^{(k)}\|_{L^p} \approx \left\|\left(\sum_{j\in\mathbb Z} 2^{2jk}|f_j|^2\right)^{1/2} \right\|_{L^p} \gtrsim (c/h)^k \left\|\left(\sum_{2^j\ge c/h} |f_j|^2\right)^{1/2} \right\|_{L^p} \tag2$$ and $$ \|f-f*\varphi_h\|_{L^p} \lesssim \left\|\left(\sum_{2^j\ge c/h} |f_j|^2\right)^{1/2} \right\|_{L^p} \tag3$$ because the low frequencies ($2^j<c/h$) cancel out in $ f-f*\varphi_h$. Comparing (2) and (3) we conclude with $\|f-f*\varphi_h\|_{L^p}\lesssim h^k \|f^{(k)}\|_{L^p}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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