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Question: Let $(X,\mathcal M, \mu)$ and $(Y, \mathcal N, \nu)$ be two $\sigma$-finite measure spaces. If functions $e_1, e_2 \ldots $ are a Hilbert space basis of $L^2(X)$ and functions $f_1, f_2, \ldots $ are a Hilbert space basis of $L^2(Y)$, prove that $e_i\cdot f_j$ is a basis of $L^2(X\times Y)$.

(Here $e_i\cdot f_j$ is a function on $X\times Y$ such that its value at $(x,y)\in X\times Y$ is $e_i(x)\cdot f_j(y)$.)

My Attempt: Assume that the basis functions are all orthonormal because if they are not, I could just use Graham Schmidt. I know that for any $x \in X$, $$x= \sum_n \langle x, e_n\rangle e_n$$ and for any $y \in Y$, $$y= \sum_m \langle y, f_m\rangle f_m$$ Now, for any element $z \in X \times Y$, I want to show that we can write it using the $e_n$ and $f_n$ basis functions. This is where I am having a tough time and was hoping to get some help. Thank you.

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    $\begingroup$ You can use the spanning criterion: Given $g \in L^2(X \times Y)$ such that $\langle g, e_i \cdot f_j \rangle = 0$ for all $i,j$, show that $g=0$. Also if your Hilbert spaces are over the complex numbers, if I'm not mistaken, you should have a complex conjugate on $f_j$, i.e. your basis should be $e_i(x) \overline{f_j}(y)$. $\endgroup$ – MisterRiemann Nov 25 '18 at 21:33
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    $\begingroup$ Thank you. You are correct about the Hilbert space conjugation in general. You wouldn’t be able to tell from my question, but I am just looking at real spaces right now. I appreciate your help. I am going to try and work that out using the spanning criterion. $\endgroup$ – MathIsHard Nov 26 '18 at 17:15
  • $\begingroup$ @MisterRiemann I am wondering how showing the spanning criterion and g=0 if the inner product is 0 for all i,j shows that the new basis works for $L^2(X x Y)$... Sorry I am just not quite seeing it. :/ $\endgroup$ – MathIsHard Nov 27 '18 at 1:15
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    $\begingroup$ For the special case $L^2$ you may want to try using Fubini's theorem $\endgroup$ – rubikscube09 Nov 27 '18 at 19:03
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    $\begingroup$ Yes, for example, see Stein and Shakarchi's Analysis Vol 3., Chapter 5(Hilbert Spaces) Question 7 $\endgroup$ – rubikscube09 Nov 27 '18 at 21:25
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To simplify things slightly, I will only present the result for separable Hilbert spaces (i.e. the ones with a countable basis, which suffices in your case).

Let $\mathcal H$ be a Hilbert space and $\{e_n\}_{n=1}^\infty \subset \mathcal H$ an orthonormal system. Then $$\{e_n\}_{n=1}^\infty \text{ is a basis for }\mathcal H \quad\iff\quad \langle x,e_n\rangle=0 \;\text{ for all } n\geq1 \text{ implies that } x = 0. $$

Since you said that you found a proof for $(\Rightarrow)$ in the literature, I will only prove $(\Leftarrow)$. So suppose that $\langle x,e_n\rangle=0$ for all $n\geq 1$ implies that $x=0$. Let $x \in \mathcal H$ and consider $$ x_n = \sum_{j=1}^n \langle x, e_j \rangle e_j. $$ For $n > m$, we then have \begin{align} \Vert x_n-x_m \Vert^2 &= \left\Vert \sum_{j=m+1}^n \langle x, e_j \rangle e_j \right\Vert^2 = \left\langle \sum_{j=m+1}^n \langle x, e_j \rangle e_j, \sum_{j=m+1}^n \langle x, e_j \rangle e_j \right\rangle\\ &= \sum_{j=m+1}^n\sum_{k=m+1}^n \langle x, e_j \rangle \overline{\langle x, e_k\rangle} \langle e_j, e_k \rangle = \sum_{j=m+1}^n |\langle x, e_j \rangle|^2, \end{align} since $\langle e_j, e_k \rangle = 0$ whenever $j\not=k$. This in fact shows that $(x_n)_{n=1}^\infty$ is a Cauchy sequence, since $$ \left(\sum_{j=1}^n |\langle x, e_j \rangle|^2\right)_{n=1}^\infty $$ is an increasing sequence which is upper bounded by $\Vert x\Vert^2$ (Bessel's inequality), and hence convergent, and in particular Cauchy. Since $\mathcal H$ is a Hilbert space, it is complete, so $x_n$ converges: $$ x_n = \sum_{j=1}^n \langle x, e_j \rangle e_j \to \tilde x = \sum_{j=1}^\infty \langle x, e_j \rangle e_j \in \mathcal H $$ It is easily checked that this limit is indeed equal to $x$, by proving that $$ \langle x-\tilde x, e_n \rangle = \lim_{k\to\infty} \langle x-x_k, e_n \rangle = 0 $$ for all $n\geq 1$, and then using the assumption. Can you do that?

Please let me know if that answers your question.

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  • $\begingroup$ Thank you very much. You are a life saver. I will make sure I understand this for the test. $\endgroup$ – MathIsHard Nov 27 '18 at 20:53
  • $\begingroup$ @MathIsHard Glad I could help. Good luck! $\endgroup$ – MisterRiemann Nov 27 '18 at 20:55

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