Orthonormal basis proof on $L_{2}[a, b]^{2}$ through Parseval's identity Assume $L_2([a,b]^{2}) = \{ f:[a,b] \times [a,b] \rightarrow \mathbb{R}: \int_{a}^{b} \int_{a}^{b} f^2(a,b)dsdt < \infty$}
with 
$\langle f, g\rangle = \int_{a}^{b}\int_{a}^{b} f(s,t)g(s,t)dsdt$, 
and assume $(\psi_k, k = 1, 2, \ldots)$ is the orthonormal basis for $L_{2}([a,b])$ space.
I'm trying to prove that $(\phi_{jk}, j,k=1,2,\ldots)$ is the orthonormal basis for $L_2([a,b]^2)$, where $\phi_{jk}$ is defined as following:
$\phi_{jk}(s,t) = \psi_j(s) \psi_{k}(t)$, $s,t \in [a,b]$. 
It is easy to show that the system is orthonormal, however, it is not clear how to prove that it is indeed a basis.
A hint here is to show that the Parseval equality holds:
$\int_{a}^{b} \int_{a}^{b} f^2(a,b)dsdt = \sum_{j,k=1}^{\infty}c^{2}_{jk}$ 
where $c_{jk}$ is defined as $c_{jk}= \int_{a}^{b}\int_{a}^{b} f(s,t) \phi_{jk}(s,t)dsdt$. 
And it is also hinted to view the problem through $f(\cdot, t): [a,b] \rightarrow \mathbb{R}$ where $t \in [a,b]$ is fixed.
My approach:
I was able to reach the hinted expression of $c_{jk}$ by going from the other side, that is, assuming that $(\phi_{jk})$ is actually an orthonormal basis, from which follows that 
$ f(s,t) = \sum_{j,k} c_{jk} \phi_{jk}(s,t)$, multiplying both sides by $\phi_{mn}(s,t)$,
$f(s,t) \phi_{mn}(s,t) = \sum_{j,k} c_{jk} \phi_{jk}(s,t) \phi_{mn}(s,t)$, integrating both sides gets
$\int_{a}^{b} \int_{a}^{b}  f(s,t) \phi_{mn}(s,t) dsdt = \sum_{j,k} c_{jk} \int_{a}^{b}  \int_{a}^{b}  \phi_{jk}(s,t) \phi_{mn}(s,t)dsdt = c_{mn} \int_{a}^{b} \int_{a}^{b} \phi^{2}_{mn}(s,t)dsdt  =c_{mn}$, where the last equations eliminates the cross-products due to orthonormality.
Similarly, multiplying the first equation by $f(s,t)$ and integrating seems to lead to the desired Parseval's equation form, that is,
$\int_{a}^{b} \int_{a}^{b} f^2(s,t)dsdt = \sum_{j,k} c_{jk} \int_{a}^{b} \int_{a}^{b} \psi_{jk}(s,t) f(s,t) dsdt = \sum_{j,k} c^2_{jk}$
however I'm not sure of the validity of such approach.
My question: What would be the correct approach for proving this? If my approach is correct, what could be the possible alternative ways of proving it by following the suggested hints? 
Also, would be grateful for any other tips or corrections! 
 A: As you write, your approach assumes the statement we want to prove. 
It's useful, however, in connecting the problem with the hint.


*

*Assume we proved $\langle f,f\rangle=\sum_{jk}\langle f,\phi_{jk}\rangle^2$ for all $f$, then no nonzero $f$ can be orthogonal to all $\phi_{jk}$'s, so they span a dense subspace.

*Actually, we can directly prove a stronger statement than Parseval's identity, namely that $f=\sum_{jk}\langle f,\phi_{jk}\rangle\phi_{jk}$. 
Let $f_t:=f(-,t)$ as suggested, check that $f_t\in L_2([a,b])$, so
$f_t(x) =\sum_j\langle f_t,\psi_j\rangle\,\psi_j(x)$ for almost all $x$,  and observe that
$$\sum_j\langle f,\phi_{jk}\rangle\phi_{jk}(x,y)\ =\  \psi_k(y)\cdot\int_a^b\psi_k(t)\cdot\left[\sum_j\psi_j(x)\cdot\int_a^bf(s,t)\psi_j(s)\,ds\right]dt=\\
=\psi_k(y)\cdot\int_a^b\psi_k(t)\cdot f_t(x)\,dt=\psi_k(y)\cdot\int_a^b\psi_k(t)f(x,t)\,dt$$
So, summing these up for $k$, for almost all $y$, we'll get the value at $y$ of the function $f(x,-)$, i.e. $f(x,y)$. 
A: Notice that $f(\cdot, t)$ is $L_2$ for almost every $t$ by Fubini's theorem. As $\{\psi_i\}$ is an orthonormal basis, we have
$$ f(\cdot, t) = \sum_j \underbrace{\left ( \int_a^b f(s, t) \psi_j(s) ds \right)}_{f_j(t)} \psi_j. $$
Now, the map $t\mapsto f_j(t)$ is again $L_2$, and we have
\begin{align*}
f_j
&= \sum_k \left ( \int_a^b  f_j(t) \psi_k(t) dt \right) \psi_k \\
&=  \sum_k \left ( \int_a^b \int_a^b f(s, t) \psi_j(s) ds \, \psi_k(t)  dt \right) \psi_k\\
&= \sum_k c_{j,k} \psi_k.
\end{align*}
And we are done. 
