2

Take $B(0,1)$ the ball in $\mathbb{R}^2$ with the normalized Lebesgue measure $\lambda$ such that $\int_{B(0,1)} d \lambda=1.$

Now, I want to show, or give a counterexample that this is false, that for all $f \in H^1_0(B(0,1))$ we have for fixed constants $a,b>0$ and any(!) $p \in (2,\infty)$ \begin{equation} ||f||_p^2 \le a \left(\int_{B(0,1)}| \nabla f|^2 d\lambda \right) + b ||f||_2^2. \end{equation}

Does anybody know how to do this? The normal Sobolev inequality is apparently too weak to show this, as this holds for any $p$ and fixed $a,b$.

user167575
  • 1,462

2 Answers2

2

If such a $a, b$ is found for a $f\in W^{1,2}_0(B)$, then $$\|f\|_p <C$$

for all $p >2$. In particular, this shows $\|f\|_\infty \le C$ as $\|f\|_p \to \|f\|_\infty$ as $p\to \infty$.

In particular, $f\in L^\infty$. Thus one cannot find such an $a, b$ for even a fixed unbounded $f\in W^{1,2}_0(B)$. In particular, pick $\phi$ be a smooth function supported in $B_{3/4}(0)$ and is $1$ on $B_{1/2}(0)$. Then

$$f(x) = -\phi(x) \log\left(\log \left(1 +\frac 1{|x|}\right)\right)$$

is such an example.

Community
  • 1
1

Can you use Sobolev embedding theorem?

Given your domain is a ball, $W_0^{1,2}(B) = H_0^{1}(B)$. In this case $W^{1,p}$ and $\mathbb{R}^n$ where $p=n=2$, we have the continuous embedding $$W^{1,2}_0(B) \hookrightarrow L^q(B) \text{ for all } q\in [1,+\infty),$$ that is $$\|f\|_q \leq C \bigg(\int_B |f|^2 dx +\int_B|\nabla f|^2 dx \bigg)^{1/2}.$$ (This is w.r.t. standard Lebesgue measure.)

Xiao
  • 9,566
  • sorry, could you state the embedding theorem you are using? This would help me quite a lot, because I feel there are 1000 different Sobolev embedding theorems in the internet. – user167575 Oct 16 '15 at 23:47
  • @user167575 see the first part of the question http://math.stackexchange.com/questions/278778/sobolev-embedding-case-p-n – Xiao Oct 16 '15 at 23:57
  • but $C$ is not the same constant for all $q$ right? – user167575 Oct 17 '15 at 00:02
  • @user167575 You are right, $C$ depends on $q$. I will take a look again. – Xiao Oct 17 '15 at 00:04