So in essence there are a few propositions in the beginning.
- For arbitrary $T$ the space $B(T;\mathbb{C})$ is a space of all (real or) complex functions $f: T\rightarrow \mathbb{C}$ with sup norm, and that is Banach space.
- Definition of $C_b(\Omega; \mathbb{C}) := B(T;\mathbb{C})\cap C(\Omega; \mathbb{C})$, which is also Banach space.
There is a reminder of weak (initial) topology on (real or complex) vector space $V$. Let there be spaces $V_\alpha$ for $\alpha \in A$ and linear operators $A_\alpha : V \rightarrow V_\alpha$. We denote the coarsest topology on set $V$, on which all defined operators are continuous (bounded), by $\tau$. $V$ with $\tau$ becomes vector space. If we use before defined linear operator to define linear operator from $V$ to product space $\Pi_{\alpha\in A} V_\alpha$ then $\tau$ is inverse image of topology inherited from product on image of that linear operator. If $V_\alpha$ are Hausdorff spaces, for $V$ with $\tau$ to be Hausdorff it is necessary and enough for the following to be valid. $$ (\forall v \in V \setminus\{0\})(\exists\alpha\in A)A_\alpha v \neq 0.$$ If spaces $V_\alpha$ are complete, and image of $V$ is closed in a product space, then $V$ is complete with $\tau$.
Now using that I have to prove two things. We have $\Omega \subseteq \mathbb{R}^{d_1+d_2}$ and set multiindex set $$\mathcal{A}:= \{\alpha = (\alpha_1, \alpha_2)\in \mathbb{N}^{d_1+d_2} : 2|\alpha_1| + |\alpha_2|\leq2 \}.$$
And the spaces which need to be proven Banach (complete) are space $$\mathrm{C}_b^{(1,2)}(\Omega ; \mathbb{C}):=\left\{u \in \mathrm{C}_b(\Omega ; \mathbb{C}):(\forall {\alpha} \in \mathcal{A}) \partial^{{\alpha}} u \in \mathrm{C}_b(\Omega ; \mathbb{C})\right\}.$$ With norm $$\|u\|:=\sum_{{\alpha} \in \mathcal{A}}\left\|\partial^{\boldsymbol{\alpha}} u\right\|_{L^{\infty}(\Omega)}.$$ And space $$\mathrm{W}^{(1,2), p}(\Omega ; \mathbb{C}):=\left\{u \in \mathrm{L}^p(\Omega ; \mathbb{C}):(\forall {\alpha} \in \mathcal{A}) \partial^{{\alpha}} u \in \mathrm{L}^p(\Omega ; \mathbb{C})\right\}.$$ With norm $$\|u\|_{\mathrm{W}^{(1,2), p}(\Omega)}:=\sqrt[p]{\sum_{{\alpha} \in \mathcal{A}}\left\|\partial^\alpha u\right\|_{\mathrm{L}^p(\Omega)}^p}.$$
Note that definition means that for $\mathrm{C}_b^{(1,2)}(\Omega ; \mathbb{C})$ all first derivations are in $C_b(\Omega; \mathbb{C})$ and second derivations only for indices $x_{d_1+1},...,x_{d_2}$.
$$u_{x_i} \in C_b(\Omega, \mathbf{C}) \quad \text{for $i = 1, 2, ... , d_1, ..., d_2$}$$ $$u_{x_ix_j} \in C_b(\Omega, \mathbf{C}) \quad \text{for $i,j = (d_1 + 1), 2, ... , d_2$}$$
Analogous for the second space. There is also need to check the details in construction on before-mentioned weak topology.
EDIT: As far as I can tell, the idea is to take space $C_b$ and linear operators $\partial^{\alpha}$ (in this case derivative in all $\mathbb{R}^{d_1+d_2}$ and second derivative only in the $d_2$ dimensions) to form two spaces whose product space is $C_b^{(1,2)}$ (and analogous for $W^{k,p}$). I am still relatively unsure how.
EDIT 2: I have found the "connected" question (and answer) in validating the construction of said topology. Proof that product topology of subspace is same as induced product topology
