## Rate of convergence of smooth mollifiers

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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?

For example, it is true that

`$$ ||u - u^{(\epsilon)}||_{L^3} \leq C \epsilon^\alpha ||u||_{B_3^{\alpha,\infty}} $$`

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).

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## 1 Answer

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.

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).

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So I see how to get the exponent.  I'm not sure what you meant about how to prove it.  (Like what is a faithfully reproduced function..?)  Are you saying that it suffices to bound $|f - f_1| \leq C ||f||$? – Phil Isett Mar 5 2011 at 14:12
"Faithfully reproduced" just means $f_\delta=f$. The usual spherical cap mollifiers reproduce constants and linear functions faithfully but have a bias on quadratic polynomials. That's why you cannot go beyond $C^2$ and $\delta^2$ with them. The way you stated it is fine too.   – fedja Mar 5 2011 at 18:26
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