# A function which belongs on a concrete Besov Space

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 $s=0$?

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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

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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 $B^{0,\infty}_\infty$.

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