If $f \in W^{-1,1}\mathbb (R^n)$,why is $\mid \hat f(\xi) \mid \leq C \mid \xi \mid$ where $C$ is a constant?

Question

If $f \in W^{-1,1}\mathbb (R^n)$, here I mean the sobolev space with negative order. Why is $\mid \hat f(\xi) \mid \leq C \mid \xi \mid$ where $C$ is a constant?

We know we can define the sobolev space of negative order by the dual space of a sobolev space, does it mean $(W^{-1,1})^* = W^{-1,\infty}$?

Answer 1

The duality-based definition of Sobolev spaces of negative order is $W^{-s, p'} = (W_0^{s,p})^*$ where $p\in [1,\infty)$ and $p'=p/(p-1)$. Notice that besides the exponent of integrability being conjugated, the order of smoothness changes sign. However, this does not help with your question because the above definition is unsuitable when $p'=1$, which would require $p=\infty$. The dual of $W^{1,\infty}_0$ is just too large to think about as a space of functions.

But one can define $W^{-1,p}(\mathbb{R}^n)$ as $\{\operatorname{div} F: F\in L^p(\mathbb{R}^n; \mathbb{R}^n)\}$ for every $p\in [1,\infty]$. Then the result you want can be obtained from the following:

• The Fourier transform of an $L^1$ function is bounded.
• Taking the partial derivative $\partial/\partial x_k$ amounts to multiplying the Fourier transform by $i\xi_k$ (maybe with $2\pi$ in there, depending on your Fourier transform convention).