figuring out two norms are equivalent

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!

• 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 4 '15 at 17:02

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.
• 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 4 '15 at 16:47
• 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 4 '15 at 17:13
• Why $||u||_{H^s}$ is comparable to $||u||_{L^2}$? I guess that's my question. – cali Nov 4 '15 at 17:26