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The following is the definition of orthonormal base that I am using:

The notion of an orthonormal basis from linear algebra generalizes over to the case of Hilbert spaces. In a Hilbert space H, an orthonormal basis is a family $\{e_{k}\}_{k} ∈ B$ of elements of $H$ satisfying the conditions:

Orthogonality: $\langle e_{k}, e_{j}\rangle = 0$ for all $k,j$ in $B$ with $k \neq j$. Normalization: $||e_{k}|| = 1$ for all $k,j$ in $B$ with $k \neq j$. Completeness: The linear span of the family $e_{k}$ with $k \in B$ is dense in $H$.

A system of vectors satisfying the first two conditions basis is called an orthonormal system or an orthonormal set. Such a system is always linearly independent. Completeness of an orthonormal system of vectors of a Hilbert space can be equivalently restated as:

if $\langle v, e_{k}\rangle = 0$ for all $k \in B$ and some $v \in H$ then $v = 0$.

This is related to the fact that the only vector orthogonal to a dense linear subspace is the zero vector, for if $S$ is any orthonormal set and $v$ is orthogonal to $S$, then $v$ is orthogonal to the closure of the linear span of $S$, which is the whole space.

In the infinite-dimensional case, an orthonormal basis will not be a basis in the sense of linear algebra; to distinguish the two, the latter basis is also called a Hamel basis. That the span of the basis vectors is dense implies that every vector in the space can be written as the sum of an infinite series, and the orthogonality implies that this decomposition is unique.

Questions:

1.How does it follow "$v$ is orthogonal to the closure of the linear span of $S$" in the explanation of the restated definition of completeness?

2.In the last paragraph it states that "The span of the basis vectors is dense in $H$ implies that every vector in the space can be written as the sum of an infinite series. How does this coincide with the usual topological definition of dense being that the closure of a subset gives you the whole space if a subset is dense. Are the two definitions equivalent?

Thanks a lot for assistance!

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A comment regarding your point 2. You say that "The span of the basis vectors is dense in $H$" implies "every vector in the space can be written as the sum of an infinite series" which is true but uses the fact that the basis vector are orthogonal. Without this requirement, you may have systems of vectors which span dense subspaces and still you cannot write every element of the ambient space as an infinite series. –  Giuseppe Negro Jan 28 at 14:53
    
The typical example is the system of all monomials $1, x, x^2, x^3...$ in the Banach space $C([0, 1])$, consisting of all continuous functions on the segment $[0,1]$ equipped with the "sup norm": $\lVert f \rVert = \sup_{x\in [0, 1]} \lvert f(x)\rvert$. The given system spans the subspace of all polynomials, which is dense (this is a famous theorem of Weierstrass). However, not all elements of $C([0,1])$ can be written as infinite series of monomials; indeed, the ones that can are precisely the analytic functions, which form a proper subclass. –  Giuseppe Negro Jan 28 at 14:55
    
Of course here we have no orthogonality: indeed, the given space is not even a Hilbert space. –  Giuseppe Negro Jan 28 at 14:56

1 Answer 1

  1. a) If $v$ is orthogonal to $S$, then it is orthogonal to any linear combination of elements of $S$ (this follows from the linearity of the scalar product), i.e. it is orthogonal to the linear span of $S$. b) If $v$ is orthogonal to some set $S$, then it is orthogonal to its closure (this follows from the continuity of the scalar product, i.e. if $x_n \to x$ and $y_n \to y$, then $\langle x_n, y_n \rangle \to \langle x, y \rangle$).

  2. The meaning of the word "dense" here is exactly the same as in the general topology.

Edit:

3. This all works in the general case of Banach spaces with Schauder bases (see, e.g., Lindenstrauss-Tzafriri's Classical Banach Spaces, vol 1, Proposition 1.a.3).

Scheme:

Define the projections $P_n(x) = \sum_{k=1}^n \langle x, e_k\rangle e_k$. Clearly, $P_n u \to u$ for all $u$ in the linear span. We want to show $P_n x \to x$ for all $x$. For that, it is enough (it's an easy argument using the density condition) to know $\|P_n\| = 1$ for all $n$. This follows from the linearity of $P_n$, $\|P_n\| \leq n$, and $\|P_n x\| \leq \|P_m x\|$ for all $n < m$ and $x$.

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Thanks for your response. One more question: How would you show that the span of the basis vectors is dense implies that every vector in the space can be written as the sum of an infinite series? I don't see how that follows. –  Moses Dec 15 '13 at 13:20
1  
Sorry, haven't been here quite a while. I edited the answer as you requested. –  ald.li Jan 28 at 14:47

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