There's some definition of concept in stein's Real analysis Page-175 made me so confused.
(0)$H$ is a Hilbert space. $T$ is a bounded symmetric linear operator on $H$
(a) a linear subspace $S$ (or simply subspace) of $H$ is a subset of $H$ that satisfies $\alpha f+\beta g \in S$ whenever $f,g \in S$ and $\alpha ,\beta$ are scalars.
(b)The subspace $S$ is closed if whenever $\{f_n \} \subset S$ converges to some $f \in H$ , then $f$ also belongs to $S$.
(c)In the case of finite-dimensional Hilbert space s,every subspace is closed.
(d)Let $S$ denote the closure of the linear space spanned by all eigenvectors of $T$ ,then $S \oplus S^\bot=H$
My confusion:
1) What does $f\in S$ and $\{f_n\}\subset S$ mean?
This question seems so obvious when $S$ defines like subspace of Riemann integrable functions in $L^2[-\pi,\pi]$. But when $S$ describes as the examples in (c),(d) , I have no idea How to deal with this question. I tried to show that the definition of $f \in H$ is : for every orthogonal basis $e_n$ of $H$ , $(f,e_n)=0$ .
But for an arbitarary subspace $S$ of $H$ , does $S$ have an orthogonal basis? If $S$ really does , then $S$ ia a Hilbert space, which leads to a contradiction.It $f$ does not have such an orthogonal basis , how to judge whether a function $f \in S$ or $\{f_n\}\subset S$ ?
2)How to prove (c)
3) For (d) , to prove $S \oplus S^\bot=H$ , I want to show $S$ is a closed subspace, But there still have some problem to deal with. What does space spanned by vectors mean.Since there might be infinitely many vectors , we can get all of the eigenvectors orthonogal , but How to prove they form a basis? In the sense that Finite linear combinations of elements in these vectors are dense in $S$.
4)How to prove (d) or how to prove $S$ is a closed subspace.
Best Answer
The notation $\{f_n\}\subset S$ means that $\{f_n\,|\,n\in\mathbb{N}\}\subset S$. In other words, you have a sequence $(f_n)_{n\in\mathbb N}$ of elements of $S$. And $S$ is closed when, whenever a sequence $(f_n)_{n\in\mathbb N}$ of elements of $S$ converges to some element $f$ of $H$, then $f\in S$.
To prove (c), let $S$ be a vector subspace of $H$ and assume that $H$ is finite-dimensional. Then there are linear forms $\alpha_1,\ldots,\alpha_k\colon H\longrightarrow\mathbb R$ such that$$S=\left\{h\in H\,\middle|\,\bigl(\forall j\in\{1,2,\ldots,k\}\bigr):\alpha_j(h)=0\right\}.$$So, if $(f_n)_{n\in\mathbb N}$ is a sequence of elements of $H$ which converges to $f\in H$ and if $j\in\{1,2,\ldots,k\}$, then $(\forall n\in\mathbb{N}):\alpha_j(f_n)=0$. Therefore$$\alpha_j(f)=\alpha_j\left(\lim_{n\to\infty}f_n\right)=\lim_{n\to\infty}\alpha_j(f_n)=\lim_{n\to\infty}0=0.$$So, $f\in S$.