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Does there exists a constant $C>0$ such that

$$ \int_{-1}^1 \lvert x\rvert\lvert\partial_x u\rvert^2 \,dx \geq C\, \lVert u\rVert^2_{H^{1/2}((-1,1))},$$ for all $u\in C^{\infty}_0((-1,1))$?

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  • $\begingroup$ Just to confirm, we have $ \| u\|^2_{H^{\frac{1}{2}}((-1,1))} := \int_{-1}^1 \frac{1}{\sqrt{1 + (2 \pi x)^2}} |\hat u(x)|^2 \, dx$? Where $\hat u (x) :=\int_{-1}^1 u(t) e^{-i x t} \, dt$. $\endgroup$
    – Nate River
    Nov 18, 2023 at 17:40
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    $\begingroup$ The integral on the frequency side must be on the whole $\mathbb R$. Additionally, the integrand factor should be inverted... so no your interpretation of the norm is incorrect $\endgroup$
    – Ali
    Nov 18, 2023 at 21:58
  • $\begingroup$ Oh sorry, made some typos. Thanks for clarifying! $\endgroup$
    – Nate River
    Nov 19, 2023 at 3:42
  • $\begingroup$ I deleted an embarrassingly nonsense answer, thanks for @an_ordinary_mathematician for spotting the mistake. $\endgroup$ Nov 19, 2023 at 16:19

1 Answer 1

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It is not true. Start with a function $u$ which vanishes for $x<0$ and is equal to $1$ for $0 \leq x \leq \frac 12$ and then smooth from $x \geq \frac 12$. The Fourier coefficients behave like $1/n$ so that $ u \not \in H^{\frac 12}$. However it can be approximated in $L^2$ with Lipschitz functions $u_\epsilon$ having $\int x |u_\epsilon'|^2 \leq C$ joining $(0,0)$ with $(\epsilon, 1)$ by a straight line. The $H^{\frac 12}$ norms of $u_\epsilon$ blow up since $u \not \in H^{\frac 12}$.

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  • $\begingroup$ However the $L^2$ norm is controlled by $\int |x| |u'|^2$. Just write (for $x>0$) $-u(x)=\int_x^1 t^{\frac 12}u'( t ) t^{-\frac 12}\, dt$ so that, by Holder, $|u(x) \leq A |\log x|^{\frac 12}$ with $A^2=\int |x||u'|^2$ $\endgroup$ Nov 21, 2023 at 18:26
  • $\begingroup$ @GeorgioMetafune It is funny that on the other hand the statement is true for even functions (simple complex interpolation). $\endgroup$
    – fedja
    Nov 27, 2023 at 19:49
  • $\begingroup$ @fedja Why is this true? I do not see it. $\endgroup$ Nov 27, 2023 at 21:57
  • $\begingroup$ Consider the linear operator $T_z:L^2\to L^2$ that maps a function $g$ to $|y|^{1-z}$ times the Fourier transform of the even extension of $G(x)=\int_x^1 g(t)t^{-z}\,dt$ from $(0,1)$ to $(-1,1)$ when $\Re z\in[0,1]$. When $\Re z=0$ and $\Re z=1$, we get a bounded mapping (in the latter case I used that $g\mapsto G$ is bounded in $L^2$, by, say, the Schur test with $\varphi(x)=\psi(x)=1/\sqrt x$). So it is bounded for $z=1/2$ as well and that is just what I said. :-) $\endgroup$
    – fedja
    Nov 27, 2023 at 22:58
  • $\begingroup$ @fedja Thank you very much, it is clear now. $\endgroup$ Nov 28, 2023 at 14:28

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