# orthonormal basis in $l^{2}$

I need an orthonormal basis in $l^{2}$. One possible choice would be to take as such the sequences $\{1,0,0,0,...\}, \{0,1,0,0,...\}, \{0,0,1,0,...\}$, but I need a basis where only finitely many components of the basis vectors are zero. Does anyone know a way to construct such a basis? One possible vector for such a basis would be $\{1,1/2,1/3,1/4,...\}$ devided by its norm. However, I don't know how to find similar vectors that are orthogonal to this one and to each other.

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Why do you want to do this? – Chris Eagle Mar 20 '12 at 13:47

How about this: consider $w = (1,1/2,1/3,1/4,\dots)$ (or any other vector with no no-zero entry), and complement it with all the standard basis vectors $e_1$, $e_2$, $\dots$ to a complete set and apply Gram-Schmidt orthonormalization?

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I already had that idea. The problem is that the other vectors that one gets this way have quite complicated entries and it is hard to do anything semi-analytical with the resulting basis. Therefore, I posted my question here in the hope that some knows a better, nice looking example. – physicist Mar 20 '12 at 11:23

You might try this: take \eqalign { x_1&\textstyle=( \rlap{1}\quad , \rlap{1\over2}\quad,\rlap{1\over3}\quad , \rlap{1\over4}\quad , \rlap{1\over5}\quad ,\ldots)\cr x_2&\textstyle=( \rlap{\alpha_1}\quad , \rlap{1\over2}\quad,\rlap{1\over3}\quad , \rlap{1\over4}\quad , \rlap{1\over5}\quad ,\ldots)\cr x_3&\textstyle=( \rlap{0}\quad , \rlap{\alpha_2}\quad,\rlap{1\over3}\quad , \rlap{1\over4}\quad , \rlap{1\over5}\quad ,\ldots)\cr x_4&\textstyle=( \rlap{0}\quad , \rlap{0}\quad,\rlap{\alpha_3}\quad , \rlap{1\over4}\quad , \rlap{1\over5}\quad ,\ldots)\cr } $$\vdots$$ for appropriately chosen scalars $\alpha_i$. Then normalize.

Note $e_1\in\text{span}\{x_1, x_2\}$, $e_2\in\text{span}\{x_1, x_2, x_3\}$, $e_3\in\text{span}\{x_1, x_2, x_3,x_4\}$, $\ldots\,$.

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Nice idea! (However, the coefficients $\alpha_k$ tend to be quite complicated.) Probably starting with the sequence $(1,1/2,1/4,1/8,\dots)$ is easier to handle. – Dirk Mar 21 '12 at 7:27

Yes, the idea of David is nice! Unfortunately, it is not what I was looking for. The problem is that the coefficients $\alpha_k$ grow to infinity when $k \rightarrow \infty$. This is the price to pay for the zeros in the beginning. Even after normalization the new coefficients, say $\widetilde{\alpha}_{k}$, will tend to $1$ and my wish was to have something that tends to zero. My initial idea was to find some permutation that does the job. For example, if I interchange the first with the second entry, the third with the fourth and so on and put a minus sign on the odd components, I get $(-\frac{1}{2},\frac{1}{1},-\frac{1}{4},\frac{1}{3},-\frac{1}{6},\frac{1}{5},...)$ which is orthogonal to $(\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},...)$ with respect to the usual scalar product in $l^2$. However, I don't know how to construct in a systematic way a third vector with these numbers that is orthogonal to the other two.

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You might want to move this as an addendum to your question, rather than as an answer (as it is not really an answer). – Arturo Magidin Mar 21 '12 at 16:04