Constrained $L^2([0,1],mathbb{R})$ space

One orthonormal basis of the subspace $M$ where $u=uchi_{(epsilon,1]}$ consists of functions that vanish on $[0,epsilon]$ and equal the following on $(epsilon,1]$: $$ c_n(x)=C_ncos(2npi (x-epsilon)/(1-epsilon)),;; nge 0, \ s_n(x)=D_nsin(2npi (x-epsilon)/(1-epsilon)),;; n ge 1. $$ The constants $C_n,D_n$ are chosen so that $|c_n|=1$ and $|d_n|=1$.

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).