Let $M:=\{ x \in \ell^{c}: \sum_{n \geq 1}x_{n}=0\}$. Note that $\ell^{c}:=\{x \in \ell^{\infty}: x_{n}=0 \operatorname{for all but finitely many} n\}$

Compute $M^{\perp}$ the orthogonal complement in $\ell^{2}$.

My ideas:

let $x \in M$ and $y \in \ell^{2}$ so that $0=\langle x,y \rangle$. Now note that $\langle x,y \rangle=\sum\limits_{n \geq 1}x_{n}y_{n}$ and since $x \in \ell ^{c}$. it immediately follows that $\sum\limits_{n \geq 1}x_{n}y_{n}=\sum\limits_{n = 1}^{N}x_{n}y_{n}$ for some particular $N$. Initially, I thought $y:=(y_{1},...,y_{N},0..)=(1,...,1,0...)$ and the span thereover would satisfy the orthogonal property, but this given $y$ of course depends on our $x$ and would therefore not be in $M^{\perp}$. Any ideas?


HINT: consider the sequence $(x_n) \in \ell^{c} $ such that $x_N = 1$, $x_{N+1} = -1$ for some $N$. Clearly, $(x_n) \in M$. Let $(y_n) \in M^{\perp}$, $(y_n)$ is perpendicular to all such $(x_n)$, where $N$ can take any values $N \geq 1$. It follows that necessarily $y_n = c$, where $c$ is a constant.

Let $c=1$. Conversely, let's show that $(y_n) \in M^{\perp}$. Take $(x_n) \in M, $ $\sum\limits_{n \geq 1}x_{n}y_{n} = \sum\limits_{n \geq 1}x_{n} = 0, $ by definition of $M$.

  • 2
    $\begingroup$ If $y_n=c$, then $c=0$ since $(y_n)\in \ell^2$. So the orthogonal complement, $M^\perp$ is trivial. $\endgroup$ – Julian Mejia Jun 3 at 21:22

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