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I have difficulty figuring out two norms are equivalent.

Let s be a nonnegatiev real number and K is a compact subset of $\mathbb{R}^d$. Let $H_{K}^s(\mathbb{R}^d)$ be the space of distributions of $H^s(\mathbb{R}^d)$ which are supported in K. Then for every $u\in H_{K}^s(\mathbb{R}^d)$, $||u||_{H^s}$ and $||u||_{\dot{H^S}}$ are equivalent.

In the book it says we only need to show $||u||_{L^2}\leq C||u||_{\dot{H^s}}$. I know how to show this, but I don't why by showing this equality we can get the answer. I only know u is compactly supported, what about $\hat{u}$? I guess probably it is not compactly supported. Then I don't know what to do. Thanks for any hint!

cali
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  • How is $\dot H^s_K$ and its norm defined? And what is this $\hat u$ you're talking about? (And what book are you referring to?) – Three.OneFour Nov 04 '15 at 17:02

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Two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent iff there is a constans $\alpha$ and $\beta$ such that for all $u$ $$\alpha \|u\|_1\leq \|u\|_2\leq\beta \|u\|_1$$ I guess the book asking only one side because the other is already discussed or trivial or so.

Michael Medvinsky
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  • Yes. One side is trivial. And the book shows the other side by showing L2 norm is smaller than $\dot{H^s}$ norm. I know why L2 is smaller than $ \dot{H^s}$ norm, but I don't know why it gives us the answer. – cali Nov 04 '15 at 16:47
  • Are you asking why this is the definition of the equivalency? Look here http://math.stackexchange.com/questions/1380191/definition-of-equivalent-norms – Michael Medvinsky Nov 04 '15 at 16:51
  • No. The L2 is not the same as the inhomogenous sobolev norm, but the book says we only need to show the L2 norm. This is where I got confused. – cali Nov 04 '15 at 17:03
  • If $\alpha |u|{H_s}\leq |u|{L_2}$ is ok with you and the book shows $|u|{L_2}\leq\beta |u|{H_s}$ then you should be convinced that they are equivalent. But I guess I didn't get your problem yet. – Michael Medvinsky Nov 04 '15 at 17:13
  • Why $||u||{H^s}$ is comparable to $||u||{L^2}$? I guess that's my question. – cali Nov 04 '15 at 17:26
  • btw, "equivalency" doesn't mean "the same", i.e. it mean the same in a very specifically defined sense. – Michael Medvinsky Nov 04 '15 at 17:27
  • most of the norms are comparable\equivalent, so this shouldn't be big surprise. – Michael Medvinsky Nov 04 '15 at 17:28
  • But I don't know how to show it... I guess it only holds when u is supported in a compact set. – cali Nov 04 '15 at 17:30
  • it may not be equivalent without this little detail (idk thought if they are or not) – Michael Medvinsky Nov 04 '15 at 17:31
  • btw, all norms on a finite dimensional space are equivalent (but this is not true for infinite dimensional spaces of course) http://math.stackexchange.com/questions/57686/understanding-of-the-theorem-that-all-norms-are-equivalent-in-finite-dimensional – Michael Medvinsky Nov 04 '15 at 17:38
  • Yes. But this case is an infinite space. I don't know how to deal with it. – cali Nov 04 '15 at 18:05
  • Two norms in an infinite space can be equivalent, just not every two. Why do you want to prove their equivalency in the case of not compactly supported function? – Michael Medvinsky Nov 04 '15 at 19:08