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I have three questions about Hilbert space that I got from Homework solution Let $H$ be Hilbert space

  1. Let $\{x_n\}_{n\ge 1}$ be a orthonormal set, but not compete. Then we can always find $y\in H,~y\ne 0$ such that $(x_n,y) = 0$ for all $n$

  2. $S$ is a subset of $H$ and its closure has non empty interior, then span$\{S\}$ is dense in $H$.

  3. If $H$ is infinite dimensional, for fixed $x\in H$, we can always find a sequence of orthonoraml $\{e_n\}_{n\ge 1}$ such that $(x,e_n)=0$ for all $n$.

user1292919
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  • What's your question? Were there parts of the solution that you didn't understand? – Jalex Stark Mar 16 '18 at 04:16
  • I stated the part that I do not understand -- those three dot points. I think it is unnecessary to state the entire questions... – user1292919 Mar 16 '18 at 04:19
  • The important omission in point (1) is that $y\ne 0.$... Point (2) is nonsense unless $H$ is $1$-dimensional. – DanielWainfleet Mar 16 '18 at 08:22
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    I highly recommend A Hilbert Space Problem Book by Paul Halmos. It is an introductory text on Hilbert space, in which many (most) basic results are presented as exercises. Solutions are included in the back of the book. – DanielWainfleet Mar 16 '18 at 08:33

1 Answers1

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There are several equivalent definitions of completeness of an orthonormal set, but I will try to answer a) using the following def.: an orthonormal set $\{e_i\}$ is complete if every vector $x$ in $H$ has an expansion $x=\sum \langle x, e_n \rangle e_n$. [The series is suppose to converge in the norm of $H$]. There is also a mistake in the statement of part 1. You should say non-zero $y$ because the zero vector is always orthogonal to each $e_n$. Suppose there is no $y \neq 0$ such that $\langle x, e_n \rangle =0$ for all $n$. Let $y=x-\sum \langle x, e_n \rangle e_n$. (You might recall that the series here is always convergent in the norm). Verify that $\langle y, e_n \rangle =0$ for all $n$ which forces $x-\sum \langle x, e_n \rangle e_n$ to be 0. This answers part 1. Prt 2. is also stated wrongly. $S \subset H$ and $\bar S$ not empty only says $S$ is non-empty and $S$ need not span $H$. (take a single vector for example). The correct statement is $S$ spans a dense subspace if $\bar S$ has non-empty interior. In this case there is an open ball $B(x,r)$ contained in $\bar S$. I leave it to you to check two facts: $B(x,r)$ spans $H$ and this implies $S$ spans a dense subspace. For Part 3 first show that there exist vectors $x_1,x_2,...$ such that $\{x_1,x_2,...\}$ is a linearly independent subset of $M\equiv \{v:\langle x, v \rangle =0\}$. This is possible because $H=M+span \{x\}$ so $M$ is also infinite dimensional. Now you have to use what is called Gram -Schmidt orhtogonalization process to get an orthonormal squenec in $M$.

Kavi Rama Murthy
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