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Let $X$ be a vector space, with a pseudo inner product space (means that it doesnt have to satisfy that $\langle u,u\rangle =0 \Leftrightarrow u=0$). Let $Y$ be a subspace of $X$. Show that the orthogonal complement of $Y$ defined as:

$$Y^{\perp}=\{x\in X:\langle x,z\rangle=0 \forall z\in Y\}$$

Show that $Y^{\perp}$ is closed.

What i've tried/done so far:

Let $(u_n)$ be a sequence of vectors from $Y^{\perp}$, which converges to a vector $u\in X$. Now if I can show that $u$ is in fact a vector in $Y^{\perp}$, then the set most the closed. So basically I want to show that $\langle u,z\rangle=0\forall z\in Y$.

I have a Lemma that states that for vectors $u,q,u_1,q_1,u_2,q_2,... \in X$ then

$$\lim_{n\rightarrow \infty}\|u_n-u\|=0=\lim_{n\rightarrow \infty}\|q_n-q\|$$

and

$$\lim_{n\rightarrow \infty}\langle u_n,q_n\rangle =\langle u,q\rangle$$

But I havent come any further, any help is much appreciated.

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For $x \in X$, define $x_\flat: X\to \Bbb R$ by $x_\flat(y)\doteq \langle x,y\rangle$. You can check that $x_\flat$ is continuous. So $x_\flat^{-1}(0)$ is closed. And $Y^\perp =\bigcap_{x\in Y} x_\flat^{-1}(0)$ is an intersection of closed sets.

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Let $S$ be any subset of $X$. For each $s\in S$, let $\phi_s$ be the bounded linear functional $\phi_s(x) = \langle x ,s\rangle $. Let $N_s$ be its null space. Then $N_s$ is a closed subspace. Furthermore

$$S^{\perp} := \{x\in X: \langle x,s\rangle =0\mbox{ for all }s\in S\} = \cap_{s\in S} N_s,$$

and therefore $S^{\perp}$ is a closed subspace.

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Let $x_{n}\in Y^{\perp}$ such that $x_{n}\rightarrow x$, then for fixed $z\in Y$, $|\left<x,z\right>|\leq|\left<x_{n}-x,z\right>|+|\left<x_{n},z\right>|\leq\|x_{n}-x\|\|z\|\rightarrow 0$, so $\left<x,z\right>=0$.

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  • $\begingroup$ Could you elaborate abit on your steps? I assume you are using Cauchy-Schwarz inequality? $\endgroup$ – HarrySames Nov 18 '17 at 20:16
  • $\begingroup$ Yes, Cauchy-Schwarz applies to $|\left<x_{n}-x,z\right>|\leq\|x_{n}-x\|\|z\|$. $\endgroup$ – user284331 Nov 18 '17 at 20:35
  • $\begingroup$ Yes, but how can you just remove $|\left<x_{n},z\right>|$, that is positive, so removing it should give someting less? $\endgroup$ – HarrySames Nov 18 '17 at 20:58
  • $\begingroup$ $|\left<x,z\right>|=|\left<x-x_{n},z\right>+\left<x_{n},z\right>|\leq|\left<x_{n}-x,z\right>|+|\left<x_{n},z\right>|$ $\endgroup$ – user284331 Nov 18 '17 at 22:25

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