Let H be a seperable Hilbert space and M a countable dense subset of H.

Show that H contains a total orthonormal sequence which can be obtained from M by the Gram-Schmidt process. (Itwasntme)

Printable View

- May 31st 2009, 01:42 PMfrater_cpGram-Schmidt Process
Let H be a seperable Hilbert space and M a countable dense subset of H.

Show that H contains a total orthonormal sequence which can be obtained from M by the Gram-Schmidt process. (Itwasntme) - May 31st 2009, 02:15 PMJose27
Let $\displaystyle B= \{e_k : k \in \mathbb{N} \}$ be your countabl dense subset of $\displaystyle H$, and $\displaystyle B_1= \{w_k : k \in \mathbb{N} \}$ the sequence obtained by applying the Gram-Schmidt process to $\displaystyle e_k$. Then let $\displaystyle w_i$ and $\displaystyle w_j$ be two elements in $\displaystyle B_1$ and $\displaystyle m=max \{i,j \}$ then applying G-S to the first $\displaystyle m$ vectors we conclude that $\displaystyle w_i \perp w_j$. So $\displaystyle B_1$ is an orthonormal set.

Now, without loss of generality, we may assume that $\displaystyle B$ is linearly independent, and since $\displaystyle B_{1,m} = \{w_k : k \leq m \}$ is contained in $\displaystyle span(B)$ for every $\displaystyle m \in \mathbb{N}$ we have $\displaystyle B_1 \subset span(B)$. Let $\displaystyle H_1 = span(B)$, since $\displaystyle B_1$ is linearly independent (actually orthogonal) and $\displaystyle \vert B_1 \vert = \vert B \vert$ we have (remember that all basis for a vector space have the same cardinality) that $\displaystyle B_1$ is a basis for $\displaystyle H_1$ and so $\displaystyle span(B_1) = H_1$. And you're finished