# Basis functions on two Hilbert spaces - showing the dot product is the basis of the product Hilbert space

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.

• 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 $e_i(x) \overline{f_j}(y)$. – MisterRiemann Nov 25 '18 at 21:33
• 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. – MathIsHard Nov 26 '18 at 17:15
• @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. :/ – MathIsHard Nov 27 '18 at 1:15
• For the special case $L^2$ you may want to try using Fubini's theorem – rubikscube09 Nov 27 '18 at 19:03
• Yes, for example, see Stein and Shakarchi's Analysis Vol 3., Chapter 5(Hilbert Spaces) Question 7 – rubikscube09 Nov 27 '18 at 21:25

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?