A question on equivalence of Sobolev norm Define $\|.\|_{T^k(\Omega)}$ as  $$ \|f\|_{T^k(\Omega)} =     \|f\|_{L^2(\Omega)} + \|(\sum\limits_{i=1}^d(\frac{\partial^{k}f}{\partial x_i^{k}})^2)^{\frac{1}{2}}\|_{L^2(\Omega)} $$
Is this norm equivalent to Sobolev norm, $\|.\|_{W^{k,2}(\Omega)}$, under two cases


*

*$\Omega$ is a bounded open subset of $\mathbb{R}^d$

*$\Omega = \mathbb{R}^d$


PS : I read that it is true, for the case $d = 1$, but I have not able to find anything for $d\ge2$ 
 A: When $k = 0,1$, the equivalence trivially holds. 
When $\Omega = \mathbb{R}^d$, the equivalence holds using the Fourier transform:
$$ \left\| \left(\sum_{i = 1}^d \left(\frac{\partial^k f}{\partial x_i^k}\right)^2\right)^{1/2} \right\|_{L^2}^2 = \sum_{i = 1}^d \left\| \frac{\partial^k f}{\partial x_i^k} \right\|_{L^2}^2 = \sum_{i = 1}^d \| (2\pi \xi_i)^k \hat{f} \|_{L^2}^2$$
Using that, by the arithmetic-geometric mean inequality, there exists a universal constant $C_k$ depending only on $k$ such that for every multiindex $\alpha$ with $|\alpha| = k$ 
$$  |\xi^\alpha| \leq C_k \sqrt{ \sum_{i = 1}^d |\xi_i|^{2k} } $$
we can run the Plancherel inequality argument in reverse to get that the $H^k$ norm is uniformly bounded by $2^{k+1} C_k$ times the $T^k$ norm. (The constant is far from optimal.) 

For general bounded open sets this can fail:
Let $\Omega$ be the following subset in $\mathbb{R}^2$:
First define $A_k = (0,2^{-2k})^2 \setminus (0,2^{-2k-1})^2$, so it looks like the letter L rotated 180 degrees. 
Set $\Omega = \cup_{k =0}^\infty A_k$ to be open. 
Define $f_k$ piecewise:
$$ f_k(x,y) = \begin{cases}
  0 & (x,y) \in A_\ell, \ell < k \\
  2^{6k} xy & (x,y) \in A_\ell, \ell \geq k.
\end{cases} $$
Due to the disjointness of the $A_\ell$, we have that $f_k$ is smooth on $\Omega$. 
One can check pretty easily (due to the self-similar scaling of $\Omega$) that 
$$ \| f_k\|_{L^2(\Omega)} = \|f_j\|_{L^2(\Omega)} $$ 
for any $j,k$. 
By definition $\partial^2_{xx} f_k = \partial^2_{yy} f_k = 0$. So $\| f_k\|_{T^2(\Omega)}$ is a constant independent of $k$. 
On the other hand, when computing $H^2$ there is the cross derivative $\partial^2_{xy} f_k$. We check that 
$$ \| \partial^2_{xy} f_k\|_{L^2(\Omega)} = \sum_{\ell = k}^\infty 2^{6k} \| 1\|_{L^2(A_\ell)} = 2^{4k} \sum_{\ell = 0}^\infty \|1\|_{L^2(A_\ell)} $$
So we see that the $H^2$ norm of $f_k$ grows like $2^{4k}$. This shows that on this particular $\Omega$ the two norms are not equivalent. 

If you do not allow arbitrary subsets, then I think $H^k$ and $T^k$ can be made equivalent if you are willing to assume, additionally, suitable regularity properties on $\partial\Omega$. Though I don't see an easy statement/proof in this direction. 
