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Suppose that $\Omega \subset \Bbb R^d$ is a sufficiently nice domain. From the examples of orthogonal bases in Hilbert space cases (or looking at a wavelets basis), it seems natural to me that one may expect the elements of a (normalized) Schauder basis $\{u_n\}_{n=1}^\infty$ of $W^{1,p}_0(\Omega)$ (for $p>1$) to be more and more "oscillatory" as $n\to\infty$. I wonder if that is true in the sense that $$ \lim_{n\to\infty} \mathscr L^d(\{ x\in\Omega : |\nabla u_n(x)| \le \varepsilon \}) = 0 $$ for any fixed $\varepsilon>0 $?

If the above is not true, is it possible to construct such a basis that satisfies the above property?

Edit: Thanks to a comment, I realized that the proposed notion of oscillation is not true (almost trivially). I want to modify it a little bit to capture what I originally had in mind. For any fixed $\varepsilon>0$, is it true that $$ \lim_{n\to\infty} \int_{\{|\nabla u_n| \le \varepsilon \}} |\nabla u_n|^p \,dx = 0 \ ? $$

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    $\begingroup$ You have your own counterexample in your question. If you take a wavelet basis, then actually $\mathcal{L}^d( \{\nabla u_n = 0\} ) \to \mathcal{L}^d(\Omega)$ as the individual wavelets are more and more localized. Though the general idea is right, you just need a better definition of oscillatory. $\endgroup$
    – mlk
    Commented Aug 21, 2021 at 7:35
  • $\begingroup$ @mlk That's true, perhaps my suggested definition was a bit naive. Is there an already established notion that captures the oscillation better and can be applied to this case? $\endgroup$
    – BigbearZzz
    Commented Aug 21, 2021 at 7:42
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    $\begingroup$ If you set $$osc_{r,x} u=\sup_{B_r(x)}(u)-\inf_{B_r(x)}(u)$$, then a classical problem sheet question is to show that if for some $\gamma\in(0,1)$, $$osc_{r,x} u < \gamma osc_{2r,x} u,$$ then $u$ is Hölder continuous. Could that give some inspiration? $\endgroup$
    – username
    Commented Aug 21, 2021 at 7:51
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    $\begingroup$ @ABIM I'd say Triebel's "Theory of function spaces" and "Function spaces and wavelets on domains" are good books to look at regarding this. I hope that helps. $\endgroup$
    – BigbearZzz
    Commented Jul 18 at 20:20
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    $\begingroup$ @ABIM I'm glad I could help ;) $\endgroup$
    – BigbearZzz
    Commented Jul 19 at 1:50

1 Answer 1

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Even the modified question does not hold.

Let $u_n$ be a basis such that $\mathcal{L}^d(\operatorname{spt} u_n) \to 0$, e.g. a wavelet basis and let $\phi \in C_0^\infty(\Omega)$ a function such that $$\int_{\{|\nabla \phi| \leq \varepsilon\}} |\nabla \phi|^p dx > 0$$ for all $\varepsilon > 0$, e.g. a smooth bump.

Now construct a modified basis $$\tilde{u}_n = a_n( \phi + u_n), $$ where the $a_n >0$ are chosen in such a way that this is again normalized. By the triangle-inequality $$ 1 = \| \tilde{u_n} \|_{W^{1,p}} \leq a_n( \| u_n \|_{W^{1,p}} + \| \phi \|_{W^{1,p}}) = a_n(1 + \| \phi \|_{W^{1,p}}), $$ so the $a_n$ are uniformly bounded from below by $a_-:= \frac{1}{1+\| \phi \|_{W^{1,p}}}$. On the other hand, since $\int_{\operatorname{spt}u_n} |\nabla \phi|^p dx \to 0$, it is easy to show that $a_n \to a_-$ and that in particular $a_n$ is bounded from above by some $a_+$.

But then $$ \int_{\{|\nabla \tilde{u}_n|\} \leq \varepsilon} |\nabla\tilde{u}_n|^p dx \geq \int_{\{|\nabla \tilde{u}_n|\leq \varepsilon\} \setminus \operatorname{spt} u_n } |\nabla\tilde{u}_n|^p dx = a_n^p \int_{\{|\nabla (a_n \phi)|\leq \varepsilon\} \setminus \operatorname{spt} u_n } |\nabla\phi|^p dx \\\geq a_-^p \int_{\{|\nabla (a_+\phi)|\leq \varepsilon\} \setminus \operatorname{spt} u_n } |\nabla\phi|^p dx \to a_-^p \int_{\{|\nabla \phi|\leq \varepsilon/a_+\}} |\nabla\phi|^p dx > 0.$$

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  • $\begingroup$ Thanks a lot! Even if the statement turns out to be false for a general Schauder basis, the wavelets basis example seems to suggest that at least it should be possible to construct a basis with the (modified) property I asked? If this is true, do you know how regular $\Omega$ should be for such a construction to be possible? $\endgroup$
    – BigbearZzz
    Commented Aug 21, 2021 at 12:47
  • $\begingroup$ @BigbearZzz If I am not fully mistaken, any wavelet-type basis would have your property. If the volume of the supports converges to 0, then the integral you gave will do so as well. $\endgroup$
    – mlk
    Commented Aug 21, 2021 at 19:50
  • $\begingroup$ I have one last question, if you don't mind. Do you happen to know where I can find a reference regarding a wavelets-like system forming a Schauder basis for $W^{1,p}_0(\Omega)$? I can easily imagine it being true for $p=2$ but I want to read more about other value of $p$. $\endgroup$
    – BigbearZzz
    Commented Aug 22, 2021 at 8:05
  • $\begingroup$ @BigbearZzz To be honest I also just assumed that such a system should be possible. At least it looks to me like a bunch of triangle functions or something similar should do it, as this is more or less what all the finite-element people use, but that is not my area of expertise. $\endgroup$
    – mlk
    Commented Aug 22, 2021 at 9:00

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