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I am reading about Sobolev spaces and I have a question regarding Sobolev spaces and the Fourier transform.

So by defining the Fourier transform, $F(\cdot)$ as an isometry we get $||f||_{L^2}=||\widehat{f}||_{L^2}$ and $\widehat{D^\alpha f}=\xi^\alpha\widehat{f}$. So one can define the norm in $H^k$, $$||f||_{H^k}=\left(\int(1+|\xi|^2)^k|\widehat{f(\xi)}|^2d\xi\right)^{\frac{1}{2}}$$ then he goes on and says that the proof that this is an equivalent norm is an easy exercise and is ommited. So I tried to prove it but I got stuck very quickly. The proof boils down to having to show that the functions $$F(\xi)=(1+|\xi|^2)^k$$ and $$S(\xi)=\sum_{|\alpha|\le k}|\xi^\alpha|^2$$ with $\xi\in \mathbb{R}^n$ bound each other. The one directions is more or less clear since $S$ includes the terms of $F$, but I cannot see why the reverse is true.

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the norm should read $||f||_{H^k}=\left(\int(1+|\xi|^2)^k|\widehat{f(\xi)}|^2d\xi\right)^{\frac{1}{2}‌​}$ – user38404 Nov 8 '12 at 16:06
I used $||\cdot||_F$ to imply that this is the norm defined by the fourier transform and not the usual Sobolev norm since I do not assume a priori that they are equivalent. – rom Nov 8 '12 at 17:15
I only meant that the $\widehat{f}$ inside the integral should be squared. – user38404 Nov 8 '12 at 17:17
Oh! I didn't notice that, you are right. Thanks. – rom Nov 8 '12 at 19:30
up vote 3 down vote accepted

You have $\xi=(\xi_1,\ldots,\xi_n)$ and $|\xi|^2=\xi_1^2+\ldots+\xi_n^2.$

If you expand $F(\xi)=(1+|\xi|^2)^k$ by binomial expansion, you will observe that the expression has precisely the same terms as $S(\xi)$. Only their coefficients differ. Therefore, you can replace the coefficients of $F(\xi)$ by a suitable number to bound $S(\xi)$ from either side.

More precisely, if $a_{\alpha}$ are the coefficients of $F(\xi)$, then a choice of the form $C_1a_{\alpha}\leq 1$ and $C_2a_{\alpha}\geq 1$ for all $\alpha$ will produce the inequalities $$C_1(1+|\xi|^2)^k\leq\sum_{|\beta|\leq k}|\xi^{\beta}|^2\leq C_2(1+|\xi|^2)^k$$

$C_1$ and $C_2$ depend on the coefficients alone and they are simply binomial coefficients which depend on $k$ and $n$.

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Interesting approach! It would be cool to see exactly how the constants deviate from each other as $k$ and $n$ vary. – Nick Alger Nov 8 '12 at 16:54
@BrianBóruma I don't mean that $F(\xi)$ and $S(\xi)$ have the same expansion, but that every power of $\xi$ that appears in $F(\xi)$ also appears in $S(\xi)$, only the coefficients of these powers differ. Could you please give an example where this is not so? – user38404 Oct 21 '14 at 6:11

There's a couple of simplifications that make it easier.

  • There is a finite number of terms in the sum $S$, so we need only bound each one individually. Ie, show that for each $\alpha \le k$, $|\xi^\alpha|^2 \le C (1+|\xi|^2)^k$.
  • We only need to consider the "tails" of the functions where $\xi$ is sufficiently large. This follows since the ratio $F/S$ is continuous in $\xi$ and therefore attains it's max and min in any compact set.
  • Since $|\xi|$ is large, we likewise need only consider the case $|\alpha|=k$.

In this regime, $$(1+|\xi|^2)^k \approx (|\xi|^2)^k = \left( \sum_i \xi_i^2 \right)^k$$

whereas $$|\xi^\alpha|^2 = \left( \sum_i \xi_i^{\alpha_i} \right)^2$$

with $\sum_i \alpha_i = k$. Applying Cauchy Schwarz with the "1-trick" yields, $$\left( \sum_i \xi_i^{\alpha_i} \right)^2 \le \left( \sum_i \xi_i^{2 \alpha_i}\right) \left(\sum_i 1^2 \right) = C \sum_i \xi_i^{2 \alpha_i}.$$

Then since $\alpha_i \le k$ and $\xi$ is sufficiently large, we have: $$\sum_i \xi_i^{2 \alpha_i} \lesssim \sum_i \xi_i^{2 k} \le \left( \sum_i \xi_i^2 \right)^k.$$

Altogether this implies $|\xi^\alpha|^2 \lesssim (1+|\xi|^2)^k$ as required.

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