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The space is $H^s(\mathbb R^d)$. If $f$ is in this space, it means

$\int_\mathbb {R^n} (1+|\xi|^2)^s|\hat f(\xi)|^2d\xi < \infty$

where $\hat f$ is the fourier transform of $f$: $\hat f(\xi)=\int_\mathbb {R^n} f(x)e^{-2\pi ix\cdot \xi}dx$.

Any good properties for this space?

I found out that if s=1, than $f$, $\nabla f\in L^2$

if s=2, then $f,\nabla f, \Delta^2f\in L^2$.

My goal is to prove

$f(x)=\frac{1}{2\pi}\int \hat f(\xi) e^{i(\xi,x)}d\xi$ where$(\xi, x)$ is the inner product=$x_1\xi_1 +x_2\xi_2...$

And I want to show that if $f\in H^s$, $|f|_{L^\infty}\leq c|f|_{H^2}$

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$H^s$ is a Sobolev space, and the statement you want to prove is the Fourier inversion formula. Both of these will be discussed at length in most graduate real analysis textbooks. Can you be more specific about what kind of "good properties" you want? – Nate Eldredge Dec 14 '12 at 19:47
I have heard that Hilbert walked into Courant's office one day and asked: “What is this thing they call Hilbert space?”. I have no idea if the story is true, but the title of your question reminded me of it. Anyhow, the Laplace operator is usually written as $\Delta$ or $\nabla^2$. What you wrote, $\Delta^2$, would be the bi-Laplacian. You'd need to be in $H^4$ to ensure that $\Delta^2f\in L^2$. – Harald Hanche-Olsen Dec 14 '12 at 19:57
But I thought the incersion formula is kind of different from this one. – Zhixia Zhang Dec 14 '12 at 20:02
up vote 0 down vote accepted

This is a Sobolev space. These spaces have many nice properties! The link there (and here) should get you going...

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