I am trying to solve the following problem:

Let $\mathcal{H}$ be a Hilbert space and $V\subset\mathcal{H}$ a closed nontrivial subspace. Let $\{e_{k}\}_{k=1}^{\infty}$ be an orthonormal basis for $\mathcal{H}$, and $P:\mathcal{H}\rightarrow V$ the orthogonal projection of $\mathcal{H}$ onto $V$. Finally, let $f_k:=Pe_k,\ k\in\mathbb{N}$.

(i) Prove that for $f\in V$,

$$f=\sum_{k=1}^{\infty}\langle f,e_k\rangle f_k.$$

(ii) Prove that despite the property (i), the family $\{f_k\}_{k=1}^{\infty}$ is not a basis for $V$. Hint: Consider any $\varphi\in V^{\perp}\setminus\{0\}$ and show that

$$\sum_{k=1}^{\infty}\langle\varphi,e_k\rangle f_k=0,$$

and use that $f=f+P\varphi$.

(iii) Argue how (i) and (ii) can be generalized to a Schauder basis $\{e_k\}_{k=1}^{\infty}$ for $\mathcal{H}$.

$\textbf{My solution:}$ Before I present what I have made so far I will skip some calculations simply because I know that they are correct and in attempt to not make this post too long.

$\textbf{Edit:}$ I have posted this before, but I have now hopefully fixed my solutions to (i) and (ii).

(i) : Since any $f\in V$ also implies $f\in\mathcal{H}$ since $V\subset\mathcal{H}$ then $f$ will have the expansion

$$f=\sum_{k=1}^{\infty}\langle f,e_k\rangle e_k.$$

Since $P$ projects $V$ onto itself this means that $Pf=f$ thus we now have

$$\begin{align*} f = Pf &= P\left(\sum_{k=1}^{\infty}\langle f,e_k\rangle e_k\right) \\ &= \sum_{k=1}^{\infty}\langle f,e_k\rangle Pe_k \\ &= \sum_{k=1}^{\infty}\langle f,e_k\rangle f_k \end{align*}$$

which is exactly what we wanted to show.

(ii) : Consider $\varphi\in V^{\perp}\setminus\{0\}$. Then we can consider any $f\in V$ as $f=f+P\varphi$, since $P\varphi=0$. With this we now have

$$\begin{align*} f=f+P\varphi &= \sum_{k=1}^{\infty}\langle f,e_k\rangle f_k + P\left(\sum_{k=1}^{\infty}\langle\varphi,e_k\rangle e_k\right) \\ &= \sum_{k=1}^{\infty}\langle f,e_k\rangle f_k + \sum_{k=1}^{\infty}\langle\varphi,e_k\rangle f_k \end{align*}$$

Since $P\varphi=0$ we now have that

$$\sum_{k=1}^{\infty}\langle\varphi,e_k\rangle f_k=0.$$

Since $\varphi\neq 0$ by construction then $\langle\varphi,e_k\rangle\neq 0$ for all $k\in\mathbb{N}$ since $e_k\neq 0$ for all $k\in\mathbb{N}$ since $\{e_k\}_{k=1}^{\infty}$ is an orthonormal basis for $\mathcal{H}$. This only leaves the option of $f_k=0$ for all $k\in\mathbb{N}$. This shows that $\{f_k\}_{k=1}^{\infty}$ can not be a basis for $V$.

$\textbf{Comment:}$ I've seen that for an orthonormal basis if $\sum_{k=1}^{\infty}\langle f,e_k\rangle f_k=0$ this should imply that $f=0$. This is why I would conclude that $\{f_k\}_{k=1}^{\infty}$ is not a basis for $V$. But what still confuses me and also makes me think I haven't done this right is the fact that $\varphi\not\in V$ so can I make the conclusion that $\{f_k\}_{k=1}^{\infty}$ is not a basis for $V$? Or can I do that since the element $f=f+P\varphi\in V$ thus $P\varphi=0\in V$, which is the sum I'm considering?

As for (iii) I have absolutely no clue, so any hints would be appreciated.

Thanks in advance!

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    $\begingroup$ the step $P\left(\sum_{k=1}^{\infty}\langle f,e_k\rangle e_k\right) = \sum_{k=1}^{\infty}\langle f,e_k\rangle Pe_k$ need some justification, because linearity just hold for finite sums. Use the continuity of $P$ for this $\endgroup$ – Masacroso Nov 14 '19 at 13:17
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    $\begingroup$ I have proven in an earlier course that this holds which is why I don't go into detail with this. $\endgroup$ – James Nov 14 '19 at 13:20
  • $\begingroup$ But do you happen to have any idea for how to solve (iii)? Cause I am really stuck on that one.. $\endgroup$ – James Nov 14 '19 at 13:26

I would argue somehow different for the end of part $(ii)$, since I don't see how You conclude $\langle\varphi,e_k\rangle\neq 0$ for all $k\in\mathbb{N}$: If $\{f_k\}_{k=1}^{\infty}$ was a basis of $V$ then from $$\sum_{k=1}^{\infty}\langle\varphi,e_k\rangle f_k=0$$ it follows $\langle\varphi,e_k\rangle=0$ for all $k\in\mathbb{N}$ by the uniqueness of representation of $0\in V$ and then since $\{e_k\}_{k=1}^{\infty}$ is a basis for $\mathcal{H}$ it follows $\varphi=0$, contradiction. Thus $\{f_k\}_{k=1}^{\infty}$ cannot be a basis of $V$.

  • $\begingroup$ I'll have to think about $(iii)$... $\endgroup$ – Peter Melech Nov 14 '19 at 13:40
  • $\begingroup$ Thank you for your answer! That makes sense. I don't know why but (iii) is for some reason really hard to grasp... $\endgroup$ – James Nov 14 '19 at 19:57

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