MathOverflow is a question and answer site for professional mathematicians. 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

Please, anyone of you know a simple example of a function which belongs to the Besov Space with $p=q=\infty$ and $s=0$ (over $\mathbb{R}$ or $I\subset\mathbb{R}$ where $I$ is a closed interval). I think that the function given by $$ \frac{1}{\log{\frac{1}{|x|^{\alpha}}}} $$ (perhaps just looking for a good $\alpha$) belongs to such space but I really don't know how to do the check.

In the other hand, and in the same way as above, some of you knows a kind of Weierstrass function, Takagi function, Minkowski question mark function,... such that it belongs to the Besov Space with $p=q=\infty$ and $s=0$?

Thanks in advance!

share|cite|improve this question
Presumably your question should include also a requirement that the function does not belong in certain other subspace of $B^{0,\infty}_\infty$, otherwise there are a lot of trivial examples. – Willie Wong Aug 31 '12 at 11:58

A Guassian $e^{-x^2}$ should work. A Guassian and it's Fourier transform decay exponentially fast. Thus, by the Fourier transform description of Besov Spaces (the only one I'm familiar with, though I understand there are other definitions), this function works, and I think is even in any Besov space

share|cite|improve this answer
Might as well just use $x\mapsto 0$.... $B^{s,p}_q$ is a vector space after all. – Willie Wong Aug 31 '12 at 11:56
ha, good point! – user7807 Aug 31 '12 at 22:16

You are talking about $B^{0,\infty}_\infty$.

Take a function $u$ in the Zygmund class $B^{1,\infty}_\infty$, which the vector space of $L^\infty$ functions such that $$\exists C,\forall x,h,\quad \vert(u(x+h)+u(x-h)-2u(x)\vert\le C\vert h\vert. $$

Note that $B^{1,\infty}_\infty\supset L^\infty\cap \text{Lipschitz}$ (here Lipschitz means $\exists C,\forall x,h,\quad \vert(u(x+h)-u(x)\vert\le C\vert h\vert $).

Now the derivative of $u$ belongs to $B^{0,\infty}_\infty$.

In particular, first derivatives of functions in $L^\infty\cap \text{Lipschitz}$ belong to $B^{0,\infty}_\infty$.

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.