Hello!

We have the following space:

$\displaystyle

Z=span \{z_r:r+2,3,...\}

$,

considered as a subspace of the Hilbert space $\displaystyle l^2$ (the space of all sequences $\displaystyle z_k$ such that $\displaystyle \sum_{k=1}^{\infty} |z_k|^2 < \infty$ , with the norm $\displaystyle \| (z_k)\|=(\sum_{k=1}^{\infty} |z_k|^2)^{\frac{1}{2}}$), where $\displaystyle z_r=(r^{-n}_{n \geq 1}\in L^2)$.

I have to find its orthogonal complement, and decide whether the space is dense and closed in $\displaystyle l^2$.

Now, here's what I got:

$\displaystyle z_r>0 \forall r$

and

$\displaystyle r_0>r_1 \rightarrow z_{r_0}<z_{r_1}$

i.e. they are strictly monotone decreasing and positive.

Hence, their linear combinations would again be either strictly decreasing or strictly increasing, and would either be all positive (in the former case) or all negative (in the latter).

Hence, we cannot approximate, say, $\displaystyle (-1,3,-10,5,4,3,2,1,0,0,0,0...) \in l^2$, by linear combinations of sequences in Z. Therefore, we have that Z is not dense in $\displaystyle l^2$.

From this (and a lemma, which says that a space is dense iff its orthogonal complement is {0}) we can conclude that the orthogonal complement is not $\displaystyle \{0\}$.

But how do I go about finding the orthogonal complement? We haven't covered orthonormal bases of Hilbert spaces yet... My intuition is the following:

$\displaystyle Z^\perp \supseteq \{(\alpha_n): \exists m_0, \exists n_0:\alpha_{m_0}>0,\alpha_{n_0}<0\} \cup \{(\alpha_n): \exists m_0,\exists n_0>m_0:\alpha_{n_0}>\alpha_{m_0}\}$

But how do I show all this rigorously??? Does the reverse inclusion hold, as well?

And I have no intuition about whether it's closed or not...