# Find a basis of $(L^2((-\pi,\pi), \mathbb{R}))^2$

I need to find a basis of $(L^2((-\pi,\pi), \mathbb{R}))^2$.

I believe a basis of $L^2((-\pi,\pi), \mathbb{R})$ can be produced by the eigenfunctions of $\triangle$ (see L.C. Evans: Partial Differential Equations, theorem 1 in section 6.5) and that one can use that as a starting point. But how proceed from there?

Or can one find a basis as eigenfunctions of $(\triangle, \triangle)^T$? What can we say about the corresponding eigenvalues?

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What is the inner product you are adopting here? – Tomás Jan 18 '13 at 14:39
@Tomás Well I thought $\langle (a,b), (c,d) \rangle_{(L^2)^2} := \langle a,c \rangle_{L^2} + \langle b,d \rangle_{L^2}$ is the natural choice. Isn't it? But the again: feel free to pick another. – mjb Jan 18 '13 at 14:47

If $\{e_n\}$ is an orthonormal basis for a Hilbert space $\mathcal{H}$ and $\{f_n\}$ is an ONB for Hilbert space $\mathcal{K}$, then the union $\{(e_n,0)\}\cup \{(0,f_n)\}$ is an ONB for the direct sum $\mathcal{H}\oplus \mathcal{K}$.

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Thank you! Can you provide a source? – mjb Jan 18 '13 at 16:26
It would likely be some exercise in a functional analysis book, but I haven't found one in the books I have. The proof is extremely straightforward: check that the pairwise inner products are what they should be (0 or 1), then notice that this systems spans $\mathcal{H}\oplus\{0\}$ and $\{0\}\oplus\mathcal{K}$. Therefore, it spans the direct sum. – user53153 Jan 18 '13 at 16:32