most recent 30 from http://mathoverflow.net 2013-05-24T20:14:43Z http://mathoverflow.net/feeds/question/57410 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57410/rate-of-convergence-of-smooth-mollifiers Phil Isett 2011-03-04T22:16:34Z 2011-03-05T01:54:11Z

<p>How does one figure out/prove the rate of convergence (in some norm) of mollifiers given a function bounded in some other norm (say Sobolev space, Besov space)? Also, is there a dimensional analysis heuristic which will predict what the rate will be?</p> <p>For example, it is true that</p> <p><code>$$ ||u - u^{(\epsilon)}||_{L^3} \leq C \epsilon^\alpha ||u||_{B_3^{\alpha,\infty}} $$</code></p> <p>where the norm on the right hand side is a Besov space norm. (This fact is used in Constantin, E and Titi's paper on Onsager's Conjecture for Solutions to Euler's Equation).</p>

http://mathoverflow.net/questions/57410/rate-of-convergence-of-smooth-mollifiers/57427#57427 fedja 2011-03-05T01:54:11Z 2011-03-05T01:54:11Z

<p>You can do something with simple scaling as long as you work on the full space. Assume that $|\cdot|$ and $\|\cdot\|$ are two shift-invariant (semi)norms and for the scaling operator $T_a f(x)=f(x/a)$ you have $|T_af|=a^t|f|$, $\|T_a f\|=a^s\|f\|$. Then, if you do the $\delta$-mollifying on $f$, it is equivalent to the $a\delta$-mollifying on $T_a f$, which means that if any inequality of the form $|f-f_\delta|\le C\delta^r\|f\|$ exists at all, you must have $a^t=a^r a^s$, i.e., $r=t-s$. Now, to check that the inequality is there, you just need to check that $\|f\|$ dominates the deviation of $f$ from some faithfully reproduced function in the norm $|\cdot|$. Note that we need the homogeneous spaces for such scaling tricks, so the Sobolev norm will really mean the $L^p$ norm of the highest derivative involved, not the full sum of $L^p$-norms of the previous derivatives.</p> <p>Example: $C$ and $C^k$ (with uniform bounds on the entire line). If we have any chance to get anything at all, the speed is $\delta^k$ by scaling. We have this chance realized if and only if our mollifier (which I assume to be compactly supported for simplicity) has first $k-1$ moments correct (i.e., reproduces the polynomials of degree up to $k-1$ precisely). </p>