21
$\begingroup$

What are the open problems in the theory of Sobolev spaces?

I would like to see problems that are yes or no only. Also I would like to see problems with the statements that are short and easy to understand for someone who has a basic knowledge in the theory, say at the level of the book by Evans and Gariepy.

The problems do not have to be a well know ones. Just the problems you think are interesting.

Please, list each problem as a separate answer.

That will allow people to leave comments related exclusively to this particular problem.

I have been working with Sobolev spaces for most of my adult live and I have some of my favorite problems that I will list below. But I will do it later, because first I would like to see your problems.

$\endgroup$
8
$\begingroup$

Let $H^{s,p}(\mathbb{R}, \mathbb{C})$ be the fractional order Sobolev space of scalar valued functions (distributions) over the real line, where $s\in \mathbb R$ and $1<p<\infty$.

It is a theorem by E. Shamir and R. Strichartz that the indicator function of the half line $1_{\mathbb{R}_+}$ (equal to $1$ for $x\geq 0$ and equal to $0$ for $x<0$) is a pointwise multiplier on $H^{s,p}(\mathbb{R}, \mathbb{C})$ if and only if ($p'$ dual exponent) $$- \frac{1}{p'} < s < \frac{1}{p}.$$ This means that $$\|1_{\mathbb{R}_+} \cdot f \|_{H^{s,p}} \leq C \|f\|_{H^{s,p}}$$ for all Schwartz functions $f$, with a constant $C > 0$ independent of $f$. This result is trivial for $s = 0$ (reducing to an $L^p$-space) but non-trivial for $s\neq 0$. Strictly outside this range, because of trace considerations, the inequality cannot hold.

My question regards the case of vector-valued functions. Let $X$ be a Banach space and let $H^{s,p}(\mathbb{R}, X)$ be the Sobolev space of $X$-valued functions (distributions), defined in the same way as in the scalar valued case. We could show the multiplier property of $1_{\mathbb{R}_+}$ in the same range as in the scalar-valued case provided the Banach space $X$ has the UMD property. See here or here, and here, Section 4 for an elementary proof of this fact. As a rule of thumb, all reflexive standard Banach spaces have UMD. Moreover, alle UMD spaces are reflexive. Space without UMD are thus $L^1$ and $L^\infty$.

My question is as follows:

Let $X$ be a Banach space. Suppose that the inequality $$\|1_{\mathbb{R}_+} \cdot f \|_{H^{s,p}(\mathbb{R}, X)} \leq C \|f\|_{H^{s,p}(\mathbb{R}, X)}$$ holds true for some $s\neq 0$ and some $1<p<\infty$, for all $X$-valued Schwartz functions $f$. Does this imply that $X$ has the UMD property?

I find this interesting because $X$ has the UMD property if and only if the Hilbert transform is a bounded operator on $L^p(\mathbb{R}, X)$, i.e. the signum function is a Fourier multiplier on this space. In other words, $F^{-1} sgn F$ is a bounded operator on $L^p(\mathbb{R}, X)$ ($F$ denoting the Fourier transform).

The pointwise multiplier property is equivalent to the boundedness of $$1_{\mathbb{R}_+} F^{-1}(1+|\cdot|^2)^{s/2} F$$ on $L^p(\mathbb{R}, X)$. So, given a positive answer the question, this would imply a new characterization of the boundedness of Hilbert transform in terms of a jump function in the time variable - and not in the frequency variable as in the usual definition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.