# Thread: a couple of hilbert spaces problems

1. ## a couple of hilbert spaces problems

(1) Let $\{e_n, n \in \mathbb{N}\}$ be an orthonormal base of the Hilbert space X. We define $Y_1$ and $Y_2$:
$Y_1=\overline{L\{e_{2n}, n \in \mathbb{N}}\}$, $Y_2=\overline{L\{e'_n=e_{2n}cos1/n + e_{2n+1}sin 1/n, n \in \mathbb{N} \}}$
Prove that $Y_1 + Y_2$ isn't closed.

I've been trying to find a convergent sequence in $Y_1 + Y_2$, with a limit outside $Y_1 + Y_2$, but no success. (Trying to prove that its complement is open would be much harder, I think.)

(2) Let $\{e_n, n \in \mathbb{N}\}$ be an orthonormal base of the Hilbert space X with the inner product $<. | .>$
We define the operator $T_n$ as follows:
$T_n x= e_1 + e_2 + e_3 + \ldots +$
$e_n + e_{n+1} + e_{n+2} + \ldots, n \in \mathbb{N}$.
Is $T_n$ bounded, normal?
Is it true that $\forall x \in X$ $T_nx \rightarrow Sx$, where S is the unilateral shift?

This one is just too messy. I can post what I've been trying to do, but didn't get anywhere and I think it wouldn't do any good.

Thank you once again for all your help.

2. Originally Posted by marianne
(1) Let $\{e_n, n \in \mathbb{N}\}$ be an orthonormal base of the Hilbert space X. We define $Y_1$ and $Y_2$:
$Y_1=\overline{L\{e_{2n}, n \in \mathbb{N}}\}$, $Y_2=\overline{L\{e'_n=e_{2n}cos1/n + e_{2n+1}sin 1/n, n \in \mathbb{N} \}}$
Prove that $Y_1 + Y_2$ isn't closed.
Hint: Let $x = \sum_{n=0}^\infty e_{2n+1}\sin 1/n$. Show that (i) this sum converges (because the sequence (sin 1/n) is square-summable); (ii) $x\notin Y_1 + Y_2$; (iii) each finite partial sum $\sum_{n=0}^N e_{2n+1}\sin 1/n$ is in Y_1+Y_2.

Originally Posted by marianne
(2) Let $\{e_n, n \in \mathbb{N}\}$ be an orthonormal base of the Hilbert space X with the inner product $<. | .>$
We define the operator $T_n$ as follows:
$T_n x= e_1 + e_2 + e_3 + \ldots +$
$e_n + e_{n+1} + e_{n+2} + \ldots, n \in \mathbb{N}$.
Is $T_n$ bounded, normal?
Is it true that $\forall x \in X$ $T_nx \rightarrow Sx$, where S is the unilateral shift?
Let Y_n be the subspace of X spanned by the first n basis vectors. Then T_n permutes these basis vectors cyclically, and it follows that the restriction of T_n to Y_n has norm 1 and is normal. On the orthogonal complement of Y_n, T_n acts as the identity. Hence T_n is bounded (with norm 1) and normal.

Clearly $T_ne_n\to Se_n$ for each basis vector. Since linear combinations of basis vectors are dense in X, and the operators T_n are uniformly bounded, it follows that $T_nx \to Sx$ for all x. [This shows that the set of normal operators is not closed, since S is not normal.]

3. Thank you very, very much!
I understood the solution to the second problem, but I'm not so sure about the first, since I don't know a lot about series.. It is clear to me why every finite sum is in fact in $Y_1 + Y_2$, but I don't know how to show that the sum converges, or that $x \notin Y_1+Y_2$.
Could you please explain a bit, or recommend an online book or something like that, that would clear it for me?

(I'm afraid I can't go to the library because I broke my leg, nor can I see my teaching asistant for this. I have this exam next week and online help is all I can get.)

4. Originally Posted by marianne
It is clear to me why every finite sum is in fact in $Y_1 + Y_2$, but I don't know how to show that the sum converges, or that $x \notin Y_1+Y_2$.
Could you please explain a bit, or recommend an online book or something like that, that would clear it for me?
Okay, here's what you need to know (I don't know of an online reference for proofs, though).

If $\{e_n\}_{n\in\mathbb{N}}$ is an orthonormal basis for a Hilbert space H, then every element of H can be expressed as an "infinite linear combination" of basis vectors. In other words, if $x\in H$ then there exist scalars $\alpha_n\;\;(n\in\mathbb{N})$ such that $\textstyle x=\sum\alpha_ne_n$, where the series converges in the sense that the partial sums converge in norm to x. The coefficients are given by the formula $\alpha_n = \langle x,e_n\rangle$. What's more, the coefficients are square-summable, with $\textstyle\sum|\alpha_n|^2 = \|x\|^2$ (Parseval's identity).

Coming back to the problem about the two subspaces, if $x = \sum_{n=0}^\infty e_{2n+1}\sin 1/n \in Y_1+Y_2$ then $x=y+z$ with $y\in Y_1,\;z\in Y_2$. But $\{e_{2n}, n \in \mathbb{N}\}$ is an orthonormal basis for Y_1, and $\{e_{2n}\cos(1/n) + e_{2n+1}\sin (1/n), n \in \mathbb{N} \}$ is an orthonormal basis for Y_2. So we can write $\textstyle y = \sum\beta_ne_{2n}$, $z=\sum\gamma_n(e_{2n}\cos(1/n) + e_{2n+1}\sin (1/n))$. Also, $\textstyle x=\sum\alpha_ne_n$. You can now compare the coefficients of the basis vectors e_n when you write the equation x=y+z using these series. You will find that this leads to the conclusion that the coefficients β_n will not be square-summable. This is a contradiction, and therefore $x\notin Y_1+Y_2.$

5. Everything makes sense now! Thank you soooooooooooooo much!