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My question is:

If $\lbrace e_n \rbrace$ is an orthnormal basis of $L^2([0,1])$, is $\lbrace {e_n}_{|(0,1)} \rbrace$ an orthonormal basis of $L^2((0,1))$?

As the points $\lbrace 1 \rbrace$ and $\lbrace 0 \rbrace$ are sets of (Lebesgue-)measure zero the scalar products on $L^2([0,1])$ and $L^2((0,1))$ take the same values for functions that agree on $(0,1)$, right? So the orthnormality should not be the problem.

If $\lbrace e_n \rbrace_{n \in \mathbb{N}}$ is a basis of $L^2([0,1])$ then for any $x \in L^2([0,1])$ it holds that $x = \sum_{n \in \mathbb{N}} \langle x, e_n \rangle e_n$.

Also any $y \in L^2((0,1))$ can be seen as the restriction to $(0,1)$ of a $x \in L^2([0,1])$, right?

Without formal proof I guess that $y = \sum_{n \in \mathbb{N}} = \langle y,{e_n}_{|(0,1)} \rangle {e_n}_{|(0,1)}$. Is that correct? This would show my assumption.

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Think about this: If $E$ has measure zero, the restriction of a function to $[0,1]\setminus E$ induces an surjective linear isometry between $L^2([0,1])$ and $L^2([0,1]\setminus E)$. Why is this, and why does it trivially prove your question? – Zach L. Feb 9 '13 at 15:34
up vote 2 down vote accepted

The spaces $L^2([0,1])$ and $L^2(]0,1[)$ are essentially the same. Indeed, the identity map $i : ]0,1[ \rightarrow [0,1]$ is an isomorphism of measurable space (i.e. bijective on a full-measure set and bimeasurable, and $i_* \mu = \mu$ where $\mu$ is the Lebesgues measure.

So yes, anthing you have on either works for the other, orthonormal basis included.

ADDENDUM : the conditions I have given are what is needed to guarantee isomorphism between $L^p$ spaces ; it should be noted that they are completely independant of topology. You would of course need other conditions to ensure isomorphism between spaces $C^0(K)$ and $C^0(K')$ for example.

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