[Math] Fractional Sobolev spaces definition

fourier analysisintuitionsobolev-spaces

Fractional Sobolev space $H^s_p(\mathbb R), s>0, 1<p<\infty$ is a space of tempered distributions $f$ that satisfy $F^{-1}((1+|\xi|^2)^{s/2} F(f)) \in L^p(\mathbb R)$.
Here, $F$ denotes the Fourier transform.

Why this definition looks like this? I mean why the function $(1+|\xi|^2)^{s/2}$ plays some role here? Why one does not take $(1+|\xi|^q)^{1/q},\ 0<q<\infty$ or just $|\xi|$, or something else ?

ADDITION: specifying the question: are there functions that one could use instead of $(1+|\xi|^2)^{s/2}$ in the defintion, that give rise to the same space?

Best Answer

Informally, $H^s_p$ should consist of functions whose derivatives of orders up to $s$ are in $L^p$.

Let's consider the case $s=1$ and $p=2$ for simplicity. We need functions $f$ such that $f,\nabla f\in L_2$. On the Fourier side, this means we want $\hat f\in L_2$ and $\xi\hat f\in L^2$. How to combine these conditions into one? Requiring $(1+|\xi|^2)^{1/2}\hat f$ to be in $L^2$ achieves that: this product majorizes both $\hat f$ and $\xi\hat f$, and has the same asymptotic at infinity as $\xi \hat f$.

We could use $1+|\xi|$ instead (in this case), but this multiplier has a major flaw: it is nonsmooth. Hence, its inverse Fourier transform is heavy-tailed (does not decay quickly at infinity). This makes it a pain to use it in a convolution (to which multiplication on Fourier side amounts).

The same smoothness issue rules out $(1+|\xi|^q)$ with general $q$. Using squared norm (quadratic polynomial, very smooth) is the natural choice. Then the outer exponent should be $s/2$ so that the multiplier is asymptotic to $|\xi|^s$ at infinity.

Multiplication by $|\xi|^s$ can be thought of as differentiating $s$ times. By itself, this multiplier leads to the homogeneous Sobolev space $\dot H_p^s$. However, in $H_p^s$ we also need to include lower order terms: otherwise, some functions that slowly decay at infinity will end up in $H^s_p$ despite not being in $L_p$. The additional $1$ in $(1+|\xi|^2)^{s/2}$ takes care of this.

And it certainly helps that the inverse Fourier transform of $(1+|\xi|^2)^{-s/2}$ (Bessel potential) is closely related to well-studied special functions (Bessel functions).

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