One way to define the sobolev-space with dominating mixed smoothness is $$H^s_{\text{mix}}(\mathbb{T}^d) :=\left\{f \in L^2(\mathbb{T}^d)\ \middle|\ \|f\|_{H^s_{\text{mix}}}^2 :=\sum_{m\in\mathbb{N}_0^d}\prod_{j=1}^{d} (1 + |m_j|^s) \langle f, b_m \rangle_{L^2}^2 < \infty\right\},$$
where $b_m$ is a basis of $L_2(\mathbb{T}^d).$
I want to show an equivalent norm is given by $$\|f\|_{H^s_{\text{mix}}(\mathbb{T}^d)}^2 = \sum_{\alpha \in \{0,s\}^d} \|D^\alpha f\|_{L^2}^2$$
I think
$$\sum_{\alpha \in \{0,s\}^d} \|D^\alpha f\|_{L^2}^2=\sum_{m\in\mathbb{N}^d}\langle f,b_m\rangle_{L_2}^2(\sum_{\alpha\in\{0,s\}}\prod_{j=1}^d|m_j|^{2\alpha})$$
is the first step.
Maybe someone can help with this.