Must a Schauder basis for $W^{1,p}_0(\Omega)$ be oscillatory? 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
$$
\lim_{n\to\infty} \int_{\{|\nabla u_n| \le \varepsilon \}} |\nabla u_n|^p \,dx = 0 \ ?
$$
 A: 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
$$ \int_{\{|\nabla \tilde{u}_n|\} \leq \varepsilon} |\nabla\tilde{u}_n|^p dx \geq \int_{\{|\nabla \tilde{u}_n|\leq \varepsilon\} \setminus \operatorname{spt} u_n }  |\nabla\tilde{u}_n|^p dx = a_n^p \int_{\{|\nabla (a_n \phi)|\leq \varepsilon\} \setminus \operatorname{spt} u_n } |\nabla\phi|^p dx \\\geq a_-^p \int_{\{|\nabla (a_+\phi)|\leq \varepsilon\} \setminus \operatorname{spt} u_n } |\nabla\phi|^p dx \to a_-^p \int_{\{|\nabla \phi|\leq \varepsilon/a_+\}} |\nabla\phi|^p dx > 0.$$
