# Hint: Computing the orthogonal Complement of $M$ in $\ell^{2}$

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