Constrained $L^2([0,1],\mathbb{R})$ space The space $L^2([0,1],\mathbb{R})$ admits an orthonormal basis:
$1, \sqrt{2} cos(2\pi n x), \sqrt{2} sin(2\pi n x), \quad n= 1,2,...$
What happens to $L^2([0,1],\mathbb{R})$ if we add a constrain, say $u(x)=0 , x \in [0,\epsilon]$ for every function $u \in L^2([0,1],\mathbb{R})$, with $\epsilon$ a small positive number?
Clearly the above trigonometric basis, can not be a basis for the new constrained space. So what is a new basis for this space ? Is the a rigorous way to determine the new basis of the constrained space ?
Also, it is clear that the dimensionality of the space is still infinite. But the space has changed, so is there any notion that can capture this change ?
Thanks a lot
 A: Disclaimer: This is in response to your comment and not an answer to the edited but rather the original question. In fact I am not yet sure about the answer to that. Nevertheless, I hope this is helpful to you.
Recall the construction of the Lebesgue space: $$ L^2([0, 1]) = \mathcal{L}^2([0,1]) / N $$where $\mathcal{L}^2([0,1])$ is the space of all square-integrable functions $f: [0,1] \to \mathbb{R}$ and $N=\{f=0 \, \text{a. e.}\}$. This means that elements in $L^2$ are equivalence classes of functions and thus, two functions are called equal if they differ only on a set of measure zero. In other words, assigning a value at a certain point is not something that's "interesting" or meaningful. You can do this if you really want to but it doesn't change much; say you want the set of all $[u] \in L^2([0,1])$ with $u(0)=0$ (where we identify $u$ as an $\mathcal{L}^2$-function and $[u]$ as the corresponding equivalence class). Then clearly this is a subset of $L^2([0,1])$. On the other hand we can realise this by choosing a representative for every $[u] \in L^2([0,1])$ such that $u(0)=0$, giving us the other subset relation. Hence they are equal and you can choose the same basis (by choosing appropriate equivalence class representatives that you need).
A: One orthonormal basis of the subspace $M$ where $u=u\chi_{(\epsilon,1]}$ consists of functions that vanish on $[0,\epsilon]$ and equal the following on $(\epsilon,1]$:
$$
       c_n(x)=C_n\cos(2n\pi (x-\epsilon)/(1-\epsilon)),\;\; n\ge 0, \\
       s_n(x)=D_n\sin(2n\pi (x-\epsilon)/(1-\epsilon)),\;\; n \ge 1.
$$
The constants $C_n,D_n$ are chosen so that $\|c_n\|=1$ and $\|d_n\|=1$.
