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Let $f\in H^m(\mathbb R^n)$. (Use the notation on wikipedia.) I am asked to prove that $$ \sum_{|\alpha|\leq m} \int |\partial^\alpha f|^2 \mathbb dx<\infty. $$ I am asked to use Parseval’s theorem to do this.

But isn't this obvious? The very definition of $H^m$ is the set of $L^2$ functions with weak derivatives up to order $m$, and those derivatives must have finite $L^2$ norm. Now $\int |\partial^\alpha f|^2 \mathbb dx$ is the square of $L^2$ norm, so it is finite. This means that the sum above $\sum_{|\alpha|\leq m} \int |\partial^\alpha f|^2 \mathbb dx$ is a sum of a finite number of terms, and each of the terms is finite.

So, the entire sum is finite. There is nothing for me to prove!

I might have understood the question wrong. Maybe a different definition of $H^m$ should be used? How can I work this out?

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There is indeed an alternative way to define $H^m(\mathbb{R^n})$: $$ H^m(\mathbb{R^n}) := \{u\in L^2(\mathbb{R^n})\ |\ (1+|\xi|^2)^{m/2}\, \widehat{u} \in L^2(\mathbb{R^n})\}. $$ Here $\widehat{u}:=\mathscr{F}_{\mathbb{R^n}}(u)$ denotes the Fourier transform of $u$. This definition is of course equivalent to the definition you mentioned; in fact this equivalence is probably what you are supposed to show.

Addition: Consider $u\in H^m(\mathbb{R^n})$ (as defined above) and let us show (with Parseval) that $\int_\mathbb{R^n} |\partial^\alpha u|^2\,dx<\infty$ for any $\alpha$ with $0\leq |\alpha|\leq m$: \begin{align*} \int_\mathbb{R^n} |\partial^\alpha u|^2\,dx &= \int_\mathbb{R^n} |\widehat{\partial^\alpha u}|^2\,dx \quad \text{(due to Parseval)}\\ &= \int_\mathbb{R^n} |\xi^\alpha\widehat{u}|^2\,dx\\ &\leq \int_\mathbb{R^n} \frac{|\xi|^{2|\alpha|}}{(1+|\xi|^2)^m}\,(1+|\xi|^2)^m|\widehat{u}|^2\,dx\\ &\leq \int_\mathbb{R^n} \,(1+|\xi|^2)^m|\widehat{u}|^2\,dx \qquad (\text{since }\frac{|\xi|^{2|\alpha|}}{(1+|\xi|^2)^m}\leq 1)\\ &< \infty \qquad\text{(by assumption)}. \end{align*}

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  • $\begingroup$ I am still a bit unsure because it appears that I still do not have much work to do. $(1+|\xi|^2)^{m/2}$ certainly dominates any power of $\xi$ less than or equal to $m$, hence we can multiply any power of $\xi$ and do inverse Fourier transform. And I don't really see how Parseval’s theorem is related. $\endgroup$ – Jethro Dec 30 '19 at 13:33
  • $\begingroup$ @Jethro: See the addition to my answer. $\endgroup$ – StarBug Dec 30 '19 at 14:07
  • $\begingroup$ Is that Parseval?! Ok, I am probably overthinking it, should it be that simple. Thank you. $\endgroup$ – Jethro Dec 30 '19 at 14:17
  • $\begingroup$ @Jethro: Yes, Parseval is simply $||u||_2=||\widehat{u}||_2$. $\endgroup$ – StarBug Dec 30 '19 at 14:30

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