Let $d\in\mathbb{N}^+$, $s\geq 0$, and consider the uniformly local Sobolev space

$$H^s_{u,loc}(\mathbb{R}^d):=\{f\in H^s_{loc}(\mathbb{R}^d)\,s.t.\,\|f\|_{H^s_{u,loc}}:=\sup_{x\in \mathbb{R}^d} \|f\|_{H^s(B_x)} < \infty\},$$

where $B_x$ is the ball of radius one centered at x. I expect that, given $0\leq s_0<s_1$, $\theta\in(0,1)$, and setting $s_{\theta}:=\theta s_1+(1-\theta)s_0$ the following interpolation inequality holds true:

$$(*)\quad\|f\|_{H^{s_{\theta}}}\leq \|f\|_{H^{s_{1}}_{u,loc}}^{\theta}\|f\|^{1-\theta}_{H^{s_{0}}} $$

Observe that, replacing $H^{s_{1}}_{u,loc}$ with $H^{s_1}$, we reduce to the classical interpolation inequality for Sobolev space.

Is estimate (*) actually true? In case, is there an explicit reference?

Thank you for any suggestions.

  • $\begingroup$ This is not true. Take for example $s_0=0$, $s_1=2$ and $s_\theta=1$; the inequality would imply that a bounded function with second derivatives in $L^p$ has a p-summable gradient, which is not true. In $1d$ (and for $p=1$) one can take $u'(x)=(1/x) \sin(\sqrt x)$ for $x \ge 1$. $\endgroup$ – Giorgio Metafune Oct 11 at 14:46
  • $\begingroup$ @ Giorgio Metafune. Thank you for the comment. It is not obvious to me why $\int_1^{\infty}|u'(x)|=+\infty$. $\endgroup$ – Capublanca Oct 11 at 16:46
  • $\begingroup$ This because $\int_1^\infty \frac{|sin (\sqrt x)|}{x}dx=\int_1^\infty \frac{|\sin t|}{t^2} 2tdt=\infty$ (change variable $x=t^2)$. However the function $u$ defined as the integral from $1$ to $x$ of the same function without the absolute value is bounded because of the oscillation of the sinus, and $u''$ is integrable. $\endgroup$ – Giorgio Metafune Oct 11 at 17:11
  • $\begingroup$ Ok thank you, a last doubt: why it's enough that $u$ is bounded? In principle you should check that $u\in L^1(1,\infty)$ $\endgroup$ – Capublanca Oct 11 at 17:28
  • $\begingroup$ Actually, it is also still not clear to me why $\int_1^{\infty}|\sin(t)|t^{-1}dt=+\infty$. Because in principle there still is some oscillation which could prevent the integral to diverge. $\endgroup$ – Capublanca Oct 11 at 17:34

Your Answer

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

Browse other questions tagged or ask your own question.