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I know that $C_{c}^{\infty}{([a,b])}$ is dense in $L^2([a,b])$.

Some common/standard orthonormal bases for $L^2([a,b])$ are the Fourier basis and the exponential basis ${(e^{inx} ,\forall n\in \mathbb{Z})}$. All elements of both of these bases are smooth functions (not with compact support however).

It is my understanding that since we have an orthonormal basis, these functions have to be dense in $L^2([a,b])$. I don't think that the converse holds though (dense set of functions does not imply an ONB). I am struggling a little bit with the intuition behind this.

Here is my question: are there any additional criteria that will turn a dense set of functions into an ONB in $L^2([a,b])$? Would smooth, periodic functions be sufficient? Any help would be appreciated!

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    Don't you usually use sequences of functions to fill out $L^2$. – 2017-02-13
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    yes! But of course, a basis is just a sequence of elements/vectors. In some cases, the sequence of sines and cosines which form a basis in $L^2$ are also smooth, and I know smooth functions are dense in $L^2$ but I don't think any arbitrary sequence of smooth functions will form a basis. I am trying to learn more about this situation. – 2017-02-13
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    The Gram Schmidt process comes to mind. – 2017-02-13
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    Gram-Schmidt would work if we had a basis, but I don't think there is a way to turn an arbitrary, dense set into a basis, and then use Gram-Schmidt to orthogonalize. Clearly, there are some dense sets which do form ONBs, as mentioned in my question, but what makes them special/different from other dense sets which do *not* form ONBs in $L^2$? – 2017-02-13

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I think you might be a bit confused on what an o.n. basis means in this case. The usual linear algebra definition of a basis if a set of linear independent vectors $\beta$ such that $span(\beta)$ is the whole vector space. With this definition your examples above are $\textit{not}$ an o.n. basis for $L^2$. In fact, no such countable basis exists. In functional analsis when we talk about basis (o.n. or otherwise) we usually mean $\textit{dense}$ basis, more specifically, a linearly independent, $\beta$, set such that $\overline{span(\beta)}$ is the whole space.

With this definition, given any dense set you can perform a version of Gram-Schmidt to get an o.n. basis.

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    Thank you! Can you give me a sketch of the proof that says you can take any arbitrary dense set in $L^2$ and turn into an ONB? Also would smooth, periodic functions be dense in $L^2$? – 2017-02-14