Let


* $H$ be a infinite dimensional, separable, complex Hilbert space,
* $\{v\_{1\_n}\}\_{n \in \mathbb{N}}$ be a sequence in $H$,
* $V\_1=\operatorname{span}\{v\_{1\_n}\}\_{n \in \mathbb{N}}$
* $U\_1=\overline{V\_1}$.


We know that $\{v\_{1\_n}\}\_{n \in \mathbb{N}}$ is a Schauder basis of $U\_1$


Let $\{v2\_n\}\_{n \in \mathbb{N}}$ be a sequence in $H$. Furthermore let,


* $V\_2=\operatorname{span}\{v\_{1\_n}\}\_{n \in \mathbb{N}}$,
* $U\_2=\overline{V\_2}$.


We know that $\{v\_{2\_n}\}\_{n \in \mathbb{N}}$ is a Schauder basis of $U\_2$ and furthermore, we know that $V\_1 \cap V\_2 = \{0\}$.


**The question**: Is it true that $U\_1 \cap U\_2 = \{0\}$.



Intersection of Hilbert spaces with Schauder basis
No, the intersection of the $U$'s might even be full. Note that it suffices to find $V$'s of countable dimension, since we can then use the Gram-Schmidt algorithm to find an orthonormal basis, which will be automatically Schauder.


Set $H=L^2([0,1])$, $V\_1$ the space generated by the $\mathbb1\_{[k/n,(k+1)/n)}$, and $V\_2$ that of the polynomials. They are dense, so the corresponding $U$'s are equal (they are precisely $H$). The intersection of the $V$'s is precisely the constants, but you can get rid of them by considering $H'\subset H$ the functions of zero mean, and $V'\_i=V\_i$.


For an explicit example of sequences, consider the basis of Haar wavelets in $H'$ and that of the non-constant Legendre polynomials.


