# Prob. 6, Sec. 3.3, in Kreyszig's Functional Analysis Book: What is $Y^\perp$ if $Y = \ldots \subset \ell^2$?

Let $$Y$$ be the subset of $$\ell^2$$ given by $$Y \colon= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N}} \in \ell^2 \ \colon \ \xi_{2n} = 0 \ \mbox{ for all } \ n \in \mathbb{N} \ \right\}.$$ Then $$Y$$ is a closed subspace of the Hilbert space $$\ell^2$$, and so $$Y$$ is also complete. What is $$Y^\perp$$?

Here is the relevant definition:

Let $$X$$ be an inner product space, and let $$M$$ be a non-empty subset of $$X$$. Then the orthogonal complement $$M^\perp$$ of $$M$$ is defined as follows: $$M \colon= \left\{ \ x \in X \ \colon \ \langle x, y \rangle = 0 \ \mbox{ for all } \ y \in M \ \right\}.$$

My effort:

By definition, \begin{align} & Y^\perp \\ &= \left\{ \ x \in \ell^2 \ \colon \ \langle x, y \rangle = 0 \ \forall y \in Y \ \right\} \\ &= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \left\langle \left( \xi_n \right)_{n \in \mathbb{N} }, \left( \eta_n \right)_{n \in \mathbb{N} } \right\rangle = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in Y \ \right\} \\ &= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty \xi_n \overline{ \eta_n} = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in Y \ \right\} \\ &= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty \xi_n \overline{ \eta_n} = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \mbox{ such that } \ \eta_{2n} = 0 \ \forall n \in \mathbb{N} \ \right\} \\ &= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty \xi_{2n-1} \overline{ \eta_{2n-1} } = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \mbox{ such that } \ \eta_{2n} = 0 \ \forall n \in \mathbb{N} \ \right\}. \end{align} My intuition is that $$Y^\perp = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N} \ \right\}.$$

Is it so? If yes, then how to prove this rigorously?

Another approach:

As $$\ell^2$$ is a Hilbert space and as $$Y$$ is a closed subspace of $$\ell^2$$, so $$\ell^2 = Y \oplus Y^\perp,$$ by Theorem 3.3-4 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig.

This means that every $$x \in \ell^2$$ has a unique representation $$x = y+z, \ \mbox{ where } \ y \in Y \ \mbox{ and } \ z \in Y^\perp.$$

Now if $$x \colon= \left( \xi_n \right)_{n \in \mathbb{N}} \in \ell^2$$, then, by definition, the series $$\sum \left\lvert \xi_n \right\rvert^2$$ of non-negative real numbers converges in $$\mathbb{R}$$.

Therefore, the sequence $$\left( \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \right)_{n \in \mathbb{N}}$$ of the partial sums of the series $$\sum \left\lvert \xi_n \right\rvert^2$$, which is also monotonically increasing, is convergent. Hence this sequence is bounded (from above). That is, for each $$n \in \mathbb{N}$$, we have $$\sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \leq \sum_{j=1}^{n+1} \left\lvert \xi_j \right\rvert^2,$$ and $$\lim_{n \to \infty} \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2$$ exists in $$\mathbb{R}$$, and
$$\lim_{n \to \infty} \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 = \sup \left\{ \ \sum_{j=1}^n \left\lvert \xi_j \right\rvert^2 \ \colon \ n \in \mathbb{N} \ \right\} < +\infty.$$

Now let $$y \colon= \left( \eta_n \right)_{n \in \mathbb{N} }$$ be ths sequence of (real or complex) numbers defined as follows: For each $$n \in \mathbb{N}$$, let $$\eta_{2n-1} \colon= \xi_{2n-1} \ \mbox{ and } \eta_{2n} \colon= 0.$$ Then the sequence $$\left( \sum_{j=1}^n \left\lvert \eta_j \right\rvert^2 \right)_{n \in \mathbb{N} }$$ of the partial sums of the series $$\sum \left\lvert \eta_j \right\rvert^2$$ is given by $$\sum_{j=1}^{2n-1} \left\lvert \eta_j \right\rvert^2 = \sum_{j=1}^n \left\lvert \xi_{2j-1} \right\rvert^2,$$ and $$\sum_{j=1}^{2n} \left\lvert \eta_j \right\rvert^2 = \sum_{j=1}^n \left\lvert \xi_{2j-1} \right\rvert^2$$ for all $$n \in \mathbb{N}$$. And, this sequence, being a sequence of sums of non-negative real numbers, is also monotonically increasing and bounded above by the sum of the convergent series $$\sum \left\lvert \xi_n \right\rvert^2$$. So the sequence $$\left( \sum_{j=1}^n \left\lvert \eta_j \right\rvert^2 \right)_{n \in \mathbb{N} }$$ is convergent, which means that the series $$\sum \left\lvert \eta_j \right\rvert^2$$ converges. Therefore, $$y \colon= \left( \eta_n \right)_{n \in \mathbb{N} }$$ is in $$\ell^2$$. So from the definition of $$Y$$ we can conclude that $$y \in Y$$.

Now let $$z \colon= \left( \zeta_n \right)_{n \in \mathbb{N}}$$ be the sequence of (real or complex) numbers defined as follows: For each $$n \in \mathbb{N}$$, let $$\zeta_{2n-1} \colon= 0 \ \mbox{ and } \ \zeta_{2n} \colon= \xi_{2n}.$$ Then $$x = y+z$$, and so $$z = x-y \in \ell^2$$, because $$\ell^2$$ is closed under addition and scalar multiplication.

Another way to show that $$z \colon= \left( \zeta_n \right)_{n \in \mathbb{N}}$$ is in $$\ell^2$$ is to show that the series $$\sum \left\lvert \zeta_n \right\rvert^2$$ of real numbers converges in $$\mathbb{R}$$.

Now the sequence $$\left( \sum_{j=1}^n \left\lvert \zeta_j \right\rvert^2 \right)_{n \in \mathbb{N} }$$ of the partial sums of the series $$\sum \left\lvert \zeta_n \right\rvert^2$$ is given by $$\sum_{j=1}^{2n-1} \left\lvert \zeta_j \right\rvert^2 = \sum_{j=1}^{2n-2} \left\lvert \xi_{2j} \right\rvert^2,$$ and $$\sum_{j=1}^{2n} \left\lvert \zeta_j \right\rvert^2 = \sum_{j=1}^{2n} \left\lvert \xi_{2j} \right\rvert^2$$ for all $$n \in \mathbb{N}$$. Thus this sequence, being a sequence of sums of non-negative real numbers, is monotonically increasing and bounded above by the sum of the series $$\sum \left\lvert \xi_n \right\rvert^2$$, and so this sequence is convergent, which implies that $$z \in \ell^2$$.

Thus $$y = \left( \eta_1, \eta_2, \eta_3, \ldots \right) = \left( \xi_1, 0, \xi_3, 0, \xi_5, 0, \ldots \right),$$ and $$z = \left( \zeta_1, \zeta_2, \zeta_3, \ldots \right) = \left( 0, \xi_2, 0, \xi_4, 0, \xi_6, \ldots \right),$$ and so $$\langle y, z \rangle = \sum_{n=1}^\infty \eta_n \overline{\zeta_n} = 0,$$ and so $$\langle y, z \rangle = 0$$ also. In fact, we can also show that $$\langle z, v \rangle = 0$$ for all $$v \in Y$$. Thus $$z \in Y^\perp$$.

But so far I have only been able to show that the subspace $$Z = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall \ n \in \mathbb{N} \ \right\}$$ is contained in $$Y^\perp$$.

Is what I have done so far correct, logically sound, and rigorous enough?

If so, then how to complete the above two arguments?

If not, then where do I lack?

P.S.:

In order to show that $$Y$$ is indeed closed in $$\ell^2$$, let us suppose that $$x \colon= \left( \xi_n \right)_{n \in \mathbb{N} }$$ is an element in the closure of $$Y$$ in $$\ell^2$$. Then, by Theorem 1.4-6 (a) in Kreyszig, there exists a sequence $$\left(y_m \right)_{m \in \mathbb{N} }$$, where $$y_m \colon= \left( \eta_{mn} \right)_{n \in \mathbb{N} }$$ for each $$m \in \mathbb{N}$$, converging to $$x$$ in $$\ell^2$$.

For each $$m \in \mathbb{N}$$, as $$y_m = \left( \eta_{mn} \right)_{n \in \mathbb{N} } \in Y$$, so we must have $$\eta_{m, 2n} = 0$$ for all $$n \in \mathbb{N}$$, that is, we must have $$\eta_{m2} = \eta_{m4} = \eta_{m6} = \cdots = 0.$$

Now let us choose an arbitrary natural number $$N$$. We show that the sequence $$\left( \eta_{mN} \right)_{m \in \mathbb{N} }$$ converges (in $$\mathbb{R}$$ or $$\mathbb{C}$$ as the case may be) to the number $$\xi_N$$.

As the sequence $$\left(y_m \right)_{m \in \mathbb{N} }$$ converges in $$\ell^2$$ to the point $$x$$, so given any real number $$\varepsilon > 0$$, there exists a natural number $$M$$ such that $$d_{\ell^2} \left( y_m, x \right) < \varepsilon$$ for all $$m \in \mathbb{N}$$ such that $$m > M$$. But we have $$x = \left( \xi_n \right)_{n \in \mathbb{N} }$$, and, for each $$m \in \mathbb{N}$$, we have $$y_m = \left( \eta_{m n} \right)_{n \in \mathbb{N} }$$. So we can conclude that $$\sqrt{ \sum_{n = 1}^\infty \left\lvert \eta_{m n} - \xi_n \right\rvert^2 } < \varepsilon$$ for all $$m \in \mathbb{N}$$ such that $$m > M$$.

And, so we obtain $$\left\lvert \eta_{mN} - \xi_N \right\rvert = \sqrt{ \left\lvert \eta_{mN} - \xi_N \right\rvert } \leq \sqrt{ \sum_{n = 1}^\infty \left\lvert \eta_{mn} - \xi_n \right\rvert^2 } < \varepsilon$$ for all $$m \in \mathbb{N}$$ such that $$m > M$$. Thus it follows that the sequence $$\left( \eta_{m N} \right)_{m \in \mathbb{N} }$$ converges to $$\xi_N$$.

Now since for each $$N \in \mathbb{N}$$, we have $$\xi_N = \lim_{m \to \infty} \eta_{m N},$$ and since for each $$N \in \mathbb{N}$$, we have $$\eta_{m, 2N} = 0$$, therefore we must have $$\xi_{2N} = 0$$ for all $$N \in \mathbb{N}$$, which shows that $$x = \left( \xi_{n} \right)_{n \in \mathbb{N} }$$ is in $$Y$$.

But this $$x$$ was an arbitrary element of the closure of $$Y$$ in $$\ell^2$$. Thus it follows that $$\mbox{Cl}(Y) \subset Y \subset \mbox{Cl}(Y)$$. Hence $$Y$$ is closed.

• Lately, it is unusual for your proofs to be insufficiently rigorous. However, they are often too long! There's a lot in your second proof that doesn't need to be there. – Omnomnomnom Jun 24 '17 at 14:09
• For example, there is no reason to talk about the monotonicity of the sequence of partial sums every time you deal with an infinite sum. It can be much more consicely explained that (as in your second proof) $y \in \ell^2$. – Omnomnomnom Jun 24 '17 at 14:21
• @Saaqib Mahmood how to prove the given space is closed – user Oct 18 '19 at 13:02
• @user686624 please have a look at my post again. I've just added a P.S. – Saaqib Mahmood Oct 18 '19 at 15:32
• @Saaqib Mahmood thank you very much – user Oct 18 '19 at 16:43

It is correct, but also making it more complicated than it has to be. You want to show that $$Z = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall \ n \in \mathbb{N} \ \right\}$$ is $Y^\perp$. As you wrote, it is clear that $Z\subset Y^\perp$. On the other hand $Z^c\subset (Y^\perp)^c$: take any $z\notin Z$ and take $n$ such that $z_{2n-1}\neq0$ (which exists by definition). Consider $e(2n-1)$, the vector with all $0$-s except at the $2n-1$-th coordinate, where it is $1$. Of course $e(2n-1)\in Y$ and $\langle z,e(2n-1)\rangle=z_{2n-1}\neq0$, so $z\notin (Y^\perp)^c$.
You have the correct intuition. We should indeed find that $$Y^\perp = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N} \ \right\}.$$ With the first approach, you've already showed that $$Y^\perp = \left\{ \ x \in \ell^2 \ \colon \ \langle x, y \rangle = 0 \ \forall y \in Y \ \right\} \\= \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \sum_{n = 1}^\infty \xi_{2n-1} \overline{ \eta_{2n-1} } = 0 \ \forall \left( \eta_n \right)_{n \in \mathbb{N} } \in \ell^2 \right\}$$ (note that because we reindexed the sum, there is no need to say $\eta_{2n} = 0$ for all $n$).
Now, let $S = \left\{ \ \left( \xi_n \right)_{n \in \mathbb{N} } \in \ell^2 \ \colon \ \xi_{2n-1} = 0 \ \forall n \in \mathbb{N} \ \right\}$. It's easy to show that $S \subset Y^\perp$: if $\xi_{2n-1} = 0$ for all $n$, then $\sum_{n = 1}^\infty \xi_{2n-1} \overline{ \eta_{2n-1} } = 0$ for any $(\eta_n)$.
To show that $Y^\perp \subset S$, take any $(\xi_n) \in Y^\perp$. define $\eta_n$ $$\eta_n = \begin{cases} \xi_n & n \text{ is even}\\ 0 & n \text{ is odd} \end{cases}$$ We then have $$\langle (\xi_n),(\eta_n) \rangle = 0 \implies\\ \sum_{n=1}^\infty |\xi_n|^2 = 0 \implies (\xi_n) \in S$$ as desired.
You might find that it's easier to do the problem, however, by first showing that the set $$\{e_{2n} : n \in \Bbb N\}$$ (as in section 2.3) forms a Schauder-basis for the space $Y$.