Consider $L^2([0,1],\mathbb{R})$.
Then, $1, \sqrt{2} cos(2 \pi j x), \sqrt{2} sin(2 \pi j x ), \quad j =1,2,...$
is a Schauder basis on $L^2([0,1], \mathbb{R})$.
I am curious, how does this generalize in the case
$L^2([0,1] \times [0,1], \mathbb{R}^2)$ ?
Thanks
As is mentioned in the commentary, let $\{(e_{i}(x)\}_{i \in \mathbb{N}}$ be the sequence that you mentioned. My affirmation is that $\mathcal{B}:=\{(e_{i}(x)e_{j}(y),0)\}_{i,j\in \mathbb{N}}\cup \{(0,e_{k}(x),e_{\ell}(y))\}_{k,\ell \in \mathbb{N}}$ is a basis for $L^{2}([0,1] \times [0,1],\mathbb{R}^{2})$. Indeed, I will prove something stronger.
I am going to use the following fact: $\{e_{\alpha}\}$ is a orthonormal (Schauder) basis (for the Hilbert space $H$) if and only if the following is true: if $f \in H$ satisfy $f \perp e_{\alpha}$ for every $\alpha$, then $f=0$.
It is clear that $\mathcal{B}$ is an orthonormal set, because if you take the inner product of some element of $\mathcal{B}_{1}=\{(e_{i}(x)e_{j}(y),0)\}_{i,j\in \mathbb{N}}$ with another from $\mathcal{B}_{2}=\{(0,e_{k}(x),e_{\ell}(y))\}_{k,\ell \in \mathbb{N}}$ you obtain zero. Also, you need to verify that $$\langle (e_{i}(x)e_{j}(y),0),(e_{m}(x)e_{n}(y),0) \rangle_{L^{2}([0,1],\mathbb{R}^{2})}=\delta_{i,m}\delta_{j,n}=\delta_{(i,m),(j,n)},$$ (and similar for the elements of $\mathcal{B}_{2}$) but this follows from the fact that $\{(e_{i}(x)\}_{i \in \mathbb{N}}$ is an orthonormal set in $L^{2}([0,1],\mathbb{R})$.
Let $f=(f_{1},f_{2})$ such that $f \perp b$ for every $b \in \mathcal{B}$. We have to show that $f \equiv 0$. By hypothesis, we have that \begin{align*} 0 &=\langle f,(e_{i}(x)e_{j}(y),0) \rangle_{L^{2}([0,1]^{2},\mathbb{R}^{2})} \\ &=\int_{[0,1]\times [0,1]}f^{1}e_{i}(x)e_{j}(y)dxdy \\ &=\int_{[0,1]}\left( \int_{[0,1]} f^{1}(x,y)e_{i}(x) dx \right) e_{j}(y) dy \\ &= \langle f_{i}^{1}(y),e_{j}(y) \rangle_{L^{2}([0,1],\mathbb{R}^{2})}, \end{align*} where $f_{i}^{1}(y):=\int_{[0,1]} f^{1}(x,y)e_{i}(x) dx$. Since $\{e_{j}\}_{j}$ is a Schauder basis for $L^{2}([0,1],\mathbb{R})$, we obtain that $f_{i}^{1}= 0$ for a.e. $y \in [0,1]$. Now we notice that $f_{i}^{1} = \langle f^{1}(x,y),e_{i} \rangle_{L^{2}([0,1],\mathbb{R})}$. Using that $\{e_{i}\}_{i}$ is a Schauder basis for $L^{2}([0,1],\mathbb{R})$, now we get that $f^{1}(x,y)= 0$ for a.e. $x,y \in [0,1]$. In a similar way, you can prove that if you work with the part of the hypothesis that says that $$0 =\langle f,(0,e_{k}(x)e_{\ell}(y)) \rangle_{L^{2}([0,1]^{2},\mathbb{R}^{2})},$$ you obtain that $f^{2}=0$ for a.e. $(x,y) \in [0,1]^{2}$. Finally, $$\|f\|_{L^{2}([0,1]^{2},\mathbb{R}^{2})}=\int_{[0,1]^{2}}[(f^{1})^{2}+(f^{2})^{2}]dxdy=0,$$ that is, $f=0$ (a.e. in $[0,1]^{2}$). By the fact that I mentioned above, this finishes the proof.
Remark: A similar proof works to find an orthonormal basis for $L^{2}(\Omega_{1} \times \cdots \times \Omega_{m},\mathbb{F}^{n})$, where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$.