# A fractional weighted Poincaré inequality

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 $$u\in C^{\infty}_0((-1,1))$$?

• 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 $\hat u (x) :=\int_{-1}^1 u(t) e^{-i x t} \, dt$. Nov 18, 2023 at 17:40
• 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
– Ali
Nov 18, 2023 at 21:58
• Oh sorry, made some typos. Thanks for clarifying! Nov 19, 2023 at 3:42
• I deleted an embarrassingly nonsense answer, thanks for @an_ordinary_mathematician for spotting the mistake. Nov 19, 2023 at 16:19

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 $$u \not \in H^{\frac 12}$$.
• 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 $A^2=\int |x||u'|^2$ Nov 21, 2023 at 18:26
• 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. :-) Nov 27, 2023 at 22:58