# Help with exercise regarding orthogonal projection

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.

• 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 – Masacroso Nov 14 '19 at 13:17
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 $$V$$.
• I'll have to think about $(iii)$... – Peter Melech Nov 14 '19 at 13:40