prove that linear span of an orthonormal set M of a Hilbert space is closed
I think i need a convergent seq in M and show that the limit belongs to span of M. but could not do it.
[Math] prove that linear span of an orthonormal set M of a hilbert space is closed
functional-analysis
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Here's my proof.
Let $A=\{A_j, j\in J\}$ be the collection of all closed linear subspace containing $S$, where $J$ is the index set. Let $B=\{B_i, i\in I\}$ be the collection of all linear subspace containing $S$, where $I$ is the index set. If $K$ is a set, we denote $\overline K$ the closure of $K$.
We know $\bigcap \limits_{i\in I} B_i$ is a linear subspace. Thus, $\overline {\bigcap \limits_{i\in I} B_i}$ is a closed linear subspace. Therefore, from the definition of $A$, we know $\overline{\bigcap \limits_{i\in I} B_i} \in A$. So, we know $\bigcap \limits_{j\in J} A_j \subseteq \overline{\bigcap \limits_{i\in I} B_i}$.
Next, we let $x\in \overline{\bigcap \limits_{i\in I} B_i}$. Then, there exists a convergent sequence $\{x_n\}$ such that $x_n\in \bigcap \limits_{i\in I} B_i, x_n\rightarrow x$. From the definitions of A and B, we know $\bigcap \limits_{i\in I} B_i \subseteq \bigcap \limits_{j\in J} A_j$. So, $x_n\in \bigcap \limits_{j\in J} A_j$. Moreover, because $\bigcap \limits_{j\in J} A_j$ is closed. Thus, $x\in\bigcap \limits_{j\in J} A_j$. Hence, $\overline{\bigcap \limits_{i\in I} B_i} \subseteq \bigcap \limits_{j\in J} A_j$.
Therefore, we have $\overline{\bigcap \limits_{i\in I} B_i} = \bigcap \limits_{j\in J} A_j$. Done.
A linear projection $P$ onto a subspace $\mathcal{M}$ has the properties that (a) the range of $P$ is $\mathcal{M}$, (b) projecting twice is the same as projecting once: $P^{2}=P$.
Orthogonal projection is something peculiar to an inner product space, and it is the same as closest point projection for a subspace. There can be many projections onto a subspace, but only one orthogonal projection. You would have seen the first examples of this in Calculus where you were asked to find the closest-point projection of a point $p$ onto a line or plane by finding a point $q$ on the line or plane such that $p-q$ is orthogonal to the given line or plane.
The orthogonal projection of a point $p$ onto a closed subspace $\mathcal{M}$ of a Hilbert space is the unique point $m\in \mathcal{M}$ such that $(p-m) \perp\mathcal{M}$. That is $(x-P_{\mathcal{M}}x) \perp \mathcal{M}$ uniquely determines $P_{\mathcal{M}}$, and this function is automatically linear. Orthogonal projection onto a subspace $\mathcal{M}$ is a closest point projection; that is, $$ \|x-m\| \ge \|x-P_{\mathcal{M}}x\|,\;\;\; m \in M, $$ with equality iff $m=P_{\mathcal{M}}x$.
For your case, the orthogonal projection $Px$ of $x$ onto the subspace spanned by $\{ e_{n}\}_{n=1}^{N}$ is the unique $y=\sum_{n}\alpha_{n}e_{n}$ such that $(x-\sum_{n}\alpha_{n}e_{n})\perp \mathcal{M}$. Equivalently, $$ (x-\sum_{n}\alpha_{n}e_{n}, e_{m})=0,\;\;\; m=1,2,3,\cdots,N, $$ or, using the orthonormality of $\{ e_{n} \}$, $$ (x,e_{m}) = \sum_{n}\alpha_{n}(e_{n},e_{m})=\alpha_{m}. $$ So the orthogonal projection $P$ onto the subspace $\mathcal{M}$ spanned by $\{ e_{n}\}_{n=1}^{N}$ is $$ Px = \sum_{n=1}^{N}(x,e_{n})e_{n}. $$ By design, one has $(x-Px)\perp\mathcal{M}$. In particular, $(x-Px)\perp Px$ because $Px\in\mathcal{M}$, which gives the orthogonal decomposition $x=Px+(I-P)x$, and $$ \|x\|^{2}=\|Px\|^{2}+\|(I-P)x\|^{2}. $$ So, both $P$ and $I-P$ are continuous. But $\mathcal{N}(I-P)=\mathcal{M}$ because $Px=x$ iff $x\in\mathcal{M}$, which guarantees that $\mathcal{M}=(I-P)^{-1}\{0\}$ is closed.
Best Answer
You cannot prove that, because it is false unless the orthonormal set is finite.
Every finite dimensional subspace of a normed space is closed, so the finite case does not depend on orthonormality.
For every infinite orthonormal set, there is a countably infinite subset $(e_1,e_2,\ldots)$, and you can show that $\sum\limits_{k=1}^\infty \dfrac{1}{k}e_k$ is in the Hilbert space, not in the linear span of the orthonormal set, but in the closure of the linear span of the orthonormal set.