Search Results

The first eigenvalue of the second derivative with Dirichlet b.c. on $(-1,1)$ is $\pi^2/4$ (with eigenfunction $\cos \frac{\pi x}{2}$) and then Poincare' inequality with optimal constant is $\|u\|_2^2 …

This is not true since a radial majorant might not be in $L^2$. A counterexample is $u(x,y)=(1+x^2+y^4)^{-\frac 12} \in H^1(\mathbb R^2)$. If $u^*$ is radial and majorizes $u$, and $r=\sqrt {x^2+y^2}$ …

I do not know a reference but the following argument gives the best constant. Consider the interval $[0,a]$ and $G(t,s)$ the Green function of $I-D^2$ with zero boundary conditions at $0,a$. If $u \in …

This is a partial (positive) answer for the convex case only but not every detail has been worked out. Let first $U$ be convex and smooth and all functions be in $C^3$ up to the boundary. Integrating …

This is true. Assume for example that $d \geq 3$. Since $(f_n)$ is bounded in $H^1$, it is bounded in $L^q$ for $2 \leq q \leq \frac{2d}{d-2}$, by Sobolev embedding. Moreover $f_n \to f$ strongly in $ …

This is a special case of embeddings for homogenuous Sobolev spaces and holds if $u \in L^1_{loc}$ with $\nabla u \in L^p$, $1 \leq p<n$ and the usual $p^*$. A proof of this (and much more) is in the …

This is a way to get an a-priori estimate, if I understood correctly the question. Multiply by $A$ and integrate by parts the left-hand-side. Then $$\int_{R^3}(A^2u^2+|\nabla A|^2)=-\int_{R^3}Au\nabla …

For $u \in H^2(I)\cap H^1_0 (I)$, $I=(1,2)$, the inequality is equivalent to $$\|(-\Delta)^{-s/2}v\|_\infty \le C \|v\|_2, \quad v \in L^2(I).$$ Let us use $$ (-\Delta)^{-s/2}v=\frac{1}{\Gamma (s/2)}\ …

Let me do for balls $\Omega_m$ and I use your Edit. Note that $\bar u_m=\frac{1}{|\Omega_m|} \int_{\Omega_m} u_m \to 0$ by H"older inequality and your assumption. Next I use Poincarè-Wirtinger inequa …

If $(T_n)$ is a sequence of uniformly bounded, linear operators in a Banach space $X$ nd $T_nx→0$ for every $x∈X$, then the convergence is uniform on a compact set $K$. Just fix $ϵ>0$ and cover $K$ wi …

Today I could check, finally. The proof I had in mind works in any dimension with $\alpha >(N-1)(1/2-1/p)$ (in your case $N=3$) which is not optimal. The optimal result with equality is proved in Theo …

Assume $\|u\|_{H^2} \leq 1$ and by H"older $$|u(x,y)|\leq \int_x^1 |u_x(t,y)|\, dt \leq \left (\int_x^1 t^2 u_x^2(t,y)\, dt\right )^{\frac 12}\left (\int_x^1 \frac{1}{t^2}\right )^{\frac 12} \leq \fra …

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 $1/n$ s …

There is no hope to gain summability without using boundary conditions. For example the function $\frac{1}{z \log z}$ is holomorphic, hence harmonic, and in $L^2$ in the disc (in the complex plane) c …

The function $u$ is not in $H^1$ (but you need $\Omega$ to be connected). Assume it is, then $u\wedge 1 =\chi_{\Omega \setminus \omega} \in H^1(\Omega)$ and its gradient is zero a.e. In fact, the grad …