Functions on Besov spaces and Hölder conditions

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Write $f_1=F_1'$ and $f_2=F_2'$. Then check $F_j\in B^{1+\alpha}_{\infty,\infty}$. The latter holds even with $\alpha=0$, because both $F_j$ are Lipschitz - they are antiderivatives of bounded functions. Hence, $f_1,f_2\in B^0_{\infty,\infty}$. The derivative operator reduces $\alpha$ by $1$ and leaves $p,q$ unchanged.

By the same reasoning, $L^\infty$ functions on a bounded domain are in $B^0_{\infty,\infty}$. See this MO answer.

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David Romero
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David Romero

Updated on August 01, 2022

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  • David Romero
    David Romero over 1 year

    Consider $f_1(x)=-1/\log(|x|)$ and $f_2(x)=1/\sqrt{-\log(|x|)}$ around $x=0$. It can be shown, just by contradiction, that the previous functions do not verify any (positive) Hölder condition: $$ |f(x)-f(y)|\leq C|x-y|^\alpha $$ with $C>0$ and $\alpha\in(0,1)$. When $\alpha\in\mathbb{R}$ the Besov spaces (with $p=q=\infty$) can be considered as the "natural extension" of the Hölder spaces (roughly speaking the set of functions that verifies a $\alpha$-Hölder condition).

    The question is, any of you know how to prove (or a hint) that $f_1$ and $f_2$ are in "negative" Besov space (that is $\alpha$-Hölder with $\alpha<0$ if it makes sense...)? They are in "different spaces" as $x\log(x)$ and $\sqrt{x}$?

    Thanks guys!

  • David Romero
    David Romero about 10 years
    Ok, thanks for the answer and if I have understood it correctly $f_j$ are only $B_{\infty,\infty}^0$ functions. But I still have a comment. The derivative reduces alpha by 1, therefore to construct a function which belongs $B_{\infty,\infty}^\alpha$, as the link that you refer, is to consider $F\in B_{\infty,\infty}^{\alpha+1}$ then $f=F'\in B_{\infty,\infty}^\alpha$, right? The problem is, how I can derive the Weierstrass function which is nowhere differentiable? To see $f$ on $B_{\infty,\infty}^\alpha$ with $\alpha<0$ consider $F'=f$ with $F$ $\alpha+1$. There are more ways to see them?
  • user98130
    user98130 about 10 years
    @DavidRomero The Weierstrass function is Hölder continuous, so you can work with it directly. Also, although it lacks a classical derivative, it has a distributional derivative, as does every locally integrable function. That derivative belongs to some negative-$\alpha$ Besov space. The elements of function spaces with negative orders of smoothness are not necessarily functions.
  • David Romero
    David Romero about 10 years
    Thanks but that's precisely the point. I'm looking for an "explicit" expression, not as derivative of a $\alpha+1$ function, of a function/measure/distribution... with negative order but not so pathological (as Dirac's delta for example). Any references? Thank's again!