Orthogonal functions with shrinking support

Question

This question is more or less a cross post of https://math.stackexchange.com/questions/1218660/orthogonal-functions-with-shrinking-support.

Let $X$ be a metric space (compact, if it helps) and let $Y$ be a closed subset of $X$. Fix a Borel measure $\mu$ on $X$. My question:

Is there an orthogonal set $\{u_n\}$ of functions in $L^2(X,\mu)$ such that the support of $u_n$ lies in a $2^{-n}$-neighborhood of $Y$?

I can construct such a set in the case $X = \mathbb{R}$ and $Y = \{0\}$ as follows. Partition the interval $[-2^{-n},2^{-n}]$ into subintervals $I_1, \ldots, I_{2^n}$ of equal length and define

$$u_n(t) = \begin{cases} 1 & t \in I_{2k+1} \\ -1 & t \in I_{2k} \\ 0 & t \notin [-2^{-n},2^{-n}] \end{cases}$$

This construction does not obviously generalize to much more general metric spaces. My only other idea was to apply Gram-Schmidt to an arbitrary sequence of functions whose supports converge to $Y$, but it is not clear how to do this while maintaining control of the supports.

Answer 1

As was pointed out by ifw, we should forbid $u_n$ from being identically $0$ (a.e.), in order to make the question interesting. Also, addressing ifw's additional observation, if we let $Y_n$ be the closed $2^{-n}$-neighborhood of $Y$, then we need to assume $L^2(Y_n,\mu|_{Y_n})$ is infinite-dimensional. (If $X$ is compact, or even just complete and separable, and $\mu$ is $\sigma$-finite, this simply means that the restriction of $\mu$ to $Y_n$ is not the sum of finitely many atoms.)

Under the above assumption on the dimension, it is easy to construct $\{u_n\}$ by induction. For each $k < n$, the inner product with $u_k$ defines a linear functional on $L^2(Y_n,\mu|_{Y_n})$. Since there are only finitely many linear functionals, and $L^2(Y_n,\mu|_{Y_n})$ is infinite-dimensional, there is some nonzero $u_n$ that is in the kernel of all of them.