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?
 A: 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*}
