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I apologize if this is more mathematical than physical, but this issue always seems to come up when I am solving physics problems.

Given a function of two variables $f(x,y)$ let us decompose it into an infinite series of orthogonal functions $\psi_n(x)$ such that $$ f(x,y) = \sum_n^\infty c_n(y)\psi_n(x) \tag{1}$$ and so $$ c_n(y) = \int_a^b \psi_n^*(x) f(x,y)\ dx\tag{2}$$

My question is simply how are we always allowed to decompose a function of two variables as the product of two variables (and then sum over varying products). Is there any sort of non-rigorous proof as to why this can always be done? I've used this fact many times, but I have never seen any explanation as to how/why we can so trivially separate them.

Emilio Pisanty
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Lone Wolf
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    Hi Lone Wolf, I've taken the liberty to edit the title to make it more descriptive, but of course feel free to roll it back. – Emilio Pisanty Jan 06 '16 at 17:05
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    You're right, this is not a physics question. The mathematically correct statement is that $L^2(\mathbb{R}\times\mathbb{R}) = L^2(\mathbb{R})\otimes L^2(\mathbb{R})$, so if you have two bases of $L^2(\mathbb{R})$ called $c_i,\psi_j$, then $L^2(\mathbb{R}^2)$ is spanned by all functions of the form $\psi_i(x)\cdot c_j(y)$, i.e. $f(x,y) = \sum_{i,j} a_{ij}c_i(x)\psi_j(y)$. So only if you have a function such that $a_{ij} = \delta_{ij}$, your equation $(1)$ will be true, but if you have the freedom to choose the $c_i$, then you'll be able to choose a basis such that $(1)$ holds. – ACuriousMind Jan 06 '16 at 17:08
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    I think it's intuitively more obvious if you think of the y variable in your equation (1) as a parameter rather than as a variable on the same level as x. So for a fixed y, f(x,y) is just a function of x, and of course you can express it as a linear combination of orthonormal basis functions with coefficients $c_n$. Viewed this way, equation (1) just asserts that you're doing this for each and every value of y, independently. – elifino Jan 06 '16 at 17:11
  • @ACuriousMind Thank you. Based on what you wrote, it makes sense then that the only functions which may not be expanded as such are those that are not square integrable. And I see you're right, I was ambiguous in writing the $c_n(t)$ but there exists functions in which the coefficients $c_n$ can be basis functions themselves. – Lone Wolf Jan 06 '16 at 17:29
  • You've actually got two things going on here. One is the separation $f(x,y) = g(x)h(y)$, and then the decomposition in to a linear combination over the basis sets. Each of those two feats gets a different justification. Partial duplicate: http://physics.stackexchange.com/q/41518/ – dmckee --- ex-moderator kitten Jan 06 '16 at 17:30
  • @elifino Thanks! This was very intuitive and smooths over any troubles I had with it. – Lone Wolf Jan 06 '16 at 17:30
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    Only after posting this was I also able to find this http://physics.stackexchange.com/q/64320/ which also answered my question – Lone Wolf Jan 06 '16 at 17:32
  • @dmckee Yes, I think that $f(x,y) = g(x)h(y)$ is the much more restrictive case, but is also relevant. – Lone Wolf Jan 06 '16 at 17:35

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