# A Hilbert basis for $L^2 ([0,1]\times[0,1])$

Let $\{f_n(x)\}$ and $\{g_n(x)\}$ be two Hilbert basis of $L^2 ([0,1])$ then $\{g_n(x)f_k(y)\}$ is a Hilbert basis for $L^2 ([0,1]\times[0,1])$.

Obs:

That is Orthogonal, and unitary is I proved with a slight use of Fubini-Tonelli.

It is only missing the density part of the span.

## 1 Answer

Denote by $\mathfrak{M}([0,1])$ the $\sigma$-algebra of Lebesgue measurable functions. The space $S([0,1])=\operatorname{span}\{1_{A\times B}:A\in\mathfrak{M}([0,1]),B\in\mathfrak{M}([0,1])\}$ is dense in $L^2([0,1]\times[0,1])$ Since $\{f_n(y)\}_{n=1}^\infty$, $\{g_n(x)\}_{n=1}^\infty$ are basis of $L^2([0,1])$ you can approximate each function in $S([0,1])$ by linear combiantions of functions $f_n(y)g_k(x)$. So $$\operatorname{Closure}\left(\operatorname{span}\{f_n(y)g_k(x)\}_{n=1,k=1}^{\infty,\infty}\right)=\operatorname{Closure}(S[0,1])=L^2([0,1]\times [0,1])$$

• I think I got it, but could you clarify a little more just the notation? – checkmath Feb 29 '12 at 23:41