1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $V$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.