# Norm in Sobolev space $H^m(\mathbb R^n)$ is finite: need I really prove this?

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?

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*}
• 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. – Jethro Dec 30 '19 at 13:33
• @Jethro: Yes, Parseval is simply $||u||_2=||\widehat{u}||_2$. – StarBug Dec 30 '19 at 14:30