How to prove $ A^{\perp} $ is a closed linear subspace?

Question

Suppose $ X $ is an inner product space and $ A\subseteq X $. I need to prove that $ A^{\perp} $ is a closed linear subspace of $ X $. Can anyone give me a idea?

Answer 1

Let $ x,y\in A^{\perp} $, $ a\in A $ and let $ \alpha ,\beta $ be two field elements of $ X $. Then $$ \langle \alpha x+\beta y,a\rangle =\alpha \langle x,a\rangle +\beta \langle y,a\rangle =0 .$$ Therefore $ \alpha x+\beta y\in A^{\perp} $ and hence $ A^{\perp} $ is a liner subspace.

To show $ A^{\perp} $ is closed, let $ (x_{n}) $ be a sequence in $ A^{\perp} $ such that $ (x_{n}) $ converges to $ x $. Observe that for all $ a\in A $, $$ 0=\langle 0,a\rangle =\lim_{n\rightarrow \infty}\langle x-x_{n},a\rangle =\langle x,a\rangle -\lim_{n\rightarrow \infty}\langle x_{n},a\rangle =\langle x,a\rangle .$$ Therefore $ x\in A^{\perp} $ and hence $ A^{\perp} $ is closed.

Answer 2

take a sequence $x_n$ in $ A^{\perp}$. such that $x_n\rightarrow x$ see that

as $\langle x_n,a\rangle=0\;\forall a\in A$ $\implies$ $\langle x,a\rangle=0\;\forall a\in A$ $\implies$ $x\in A^{\perp}$ and $A^{\perp}$ is closed

And you can verify $A^{\perp}$ is subspace same way

Answer 3

Let be a point $a\in X$: $$\{a\}^{\perp}=\{x\in X\,\vert\,\langle a,x\rangle=0\}$$ is closed (inverse image of the closed set $\{0\}$ by a continuous function). And $$A^{\perp}=\bigcap_{a\in A}\{a\}^{\perp}$$ is an intersection of closed sets.

Answer 4

Notation: We use $Y\subseteq X$ instead.

Show that $Y^\perp$ is a closed linear subspace of $X'$.

Let $\alpha,\beta\in\mathbb{C}$, $l_1,l_2\in Y^\perp$, $y\in Y$.

$(\alpha l_1+\beta l_2)(y)=\alpha l_1(y)+\beta l_2(y)=0$. Therefore $Y^\perp$ is linear.

Let $(l_n)$ be a sequence in $Y^\perp$ converging to $l\in X'$, i.e.\ $\|l_n-l\|\to0$. There exists $N$ such that for all $n\geq N$, $\|l_n-l\|=\sup_{x\leq 1}|(l_n-l)(x)|<\epsilon$.

Let $y\in Y$. For $n\geq N$, we have \begin{align*} |l(y)|&=\|y\||l(\frac{y}{\|y\|})|\\ &\leq\|y\||(l-l_n)(\frac{y}{\|y\|})|+\|y\||l_n(\frac{y}{\|y\|})|\\ &<\|y\|\epsilon. \end{align*} Since $\epsilon$ is arbitrary, $l(y)=0$ for all $y\in Y$. Thus $l\in Y^\perp$.