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Let $H$ be the space of measurable functions on $(0,1)$ such that $$ \|u\|_{H}^2 = \int_0^1 x^2\,|\partial_x u|^2\,dx + \int_{0}^1 |u|^2\,dx <\infty.$$

Let $C>0$ be a constant. Suppose that $W \subset H$ is a subspace such that for all $u\in H$ there holds: $$ \int_0^1 x^2\,|\partial_x u|^2\,dx \leq C \int_0^1 |u|^2\,dx.$$

Let $\delta \in (0,1)$. Is it true that the restriction set $W|_{(\delta,1)}:=\{ u|_{(\delta,1)}\,:\, u\in W\}$ is a finite dimensional subspace?

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  • $\begingroup$ I'm not sure whether I understood your question correctly but it seems to me that for a given $C>0$, all the functions $x^{-\alpha}$ with $\alpha < \min\{\frac{1}{2},\sqrt{C}\}$ belong to $W$. These functions, when restricted to any interval, are still linearly independent. $\endgroup$
    – T. Le
    Commented Oct 27, 2023 at 23:59
  • $\begingroup$ But do all of their linear combinations also satisfy the same bound? $\endgroup$
    – Ali
    Commented Oct 28, 2023 at 0:36
  • $\begingroup$ No. Let $v_n$ be a sequence of functions supported in $(0, \delta)$, with disjoint support and such that $\int |v_n|^2 dx = \int x^2 |v'_n|^2 dx=1$ (carefully chosen bump functions). Let $w_n$ be any sequence supported in $(\delta,1)$ in the unit ball of $H$. Let finally $u_n=v_n + 2^{-n} w_n$. Then the linear span $W$ of the $u_n$'s satisfy your condition, but $W\left|_{(\delta,1)}\right.$ is the span of the $w_n$'s, so it can be any countable dimension subspace. $\endgroup$
    – Mikael de la Salle
    Commented Oct 28, 2023 at 3:07
  • $\begingroup$ @Ali, I don't know whether the linear combinations of my functions also satisfy the same condition. That was what I missed when thinking about the problem! $\endgroup$
    – T. Le
    Commented Oct 28, 2023 at 13:08
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    $\begingroup$ @T.Le Yes, the factor $x^2$ is precisely what what the construction possible. Let $f$ be any compactly supported function on $(0,1)$ such that $\int |f|^2 = \int x^2 |f'_n|^2 = 1$. Then the same is true for the function $\varepsilon^{-1/2} f(\cdot/\varepsilon)$, and if the sequence $\varepsilon(n)$ decreases fast enough to $0$, then the functions $v_n=f_{\varepsilon(n)}$ have disjoint support, contained in $(0,\delta)$. $\endgroup$
    – Mikael de la Salle
    Commented Oct 30, 2023 at 13:41

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