1
$\begingroup$

Let $V$ be a complete subspace of $\ell^2$ over the complex plane $\mathbb{C}$

Let $T:V \to \mathbb{C}$ be a bounded linear operator

Let $w \in V$ such that $T(v)=\langle v,w\rangle $ (Riesz representation theorem)

Let $u \in \ell^2 \backslash V $ be a vector of $\ell^2$ but not in $V$

I would like to know if is true that $$ \langle u,w\rangle =0 $$

Thanks.

$\endgroup$
  • $\begingroup$ If $V$ is a nontrivial proper subspace and $T$ is not identically zero, then the answer is no because $u\in \ell ^2\setminus V$ does not imply $u\in V^\perp$. See my post for details. $\endgroup$ – Pedro May 31 '17 at 17:09
1
$\begingroup$

It is not true in general. For example consider $e_1 = (1,0,0,\dots)$ and let $V = \text{span}\{e_1\}$. Then $V$ is a complete subset of $\ell^2$, but $e_1 + e_2 \notin V$ and $\langle e_1 + e_2,e_1\rangle = 1 \neq 0$.

As you can see, here $T$ doesn't really play any role, but if you want you can think of $e_1$ as if it was given by the linear map $T(v) = \langle v,e_1\rangle$.

$\endgroup$
  • $\begingroup$ thanks so much @Giovanni $\endgroup$ – Matey Math May 31 '17 at 16:14
  • $\begingroup$ @MateyMath: you are quite welcome! $\endgroup$ – Giovanni May 31 '17 at 16:27
  • $\begingroup$ @MateyMath and Giovanni Actually, always it is not true (provided that $V$ is a nontrivial proper subspace and $T$ is not identically zero, as in this counterexample). And the particular "sequence structure" of $\ell^2$ also does not play any important role (tha same is valid for any Hilbert space). See my post. $\endgroup$ – Pedro May 31 '17 at 17:10
  • $\begingroup$ @Pedro: yes, of course, and the argument is exactly the same. $\endgroup$ – Giovanni May 31 '17 at 17:18
2
$\begingroup$

If $V=\{0\}$ or $T$ is identically null, then the result holds trivially (because in these cases we have $w=0$). So, let us assume $V\neq \{0\}$ and $T\neq 0$. Let us also assume $V\neq \ell^2$ to be possible to take $u\in \ell^2\setminus V$.

As $V$ is closed, we have $\ell^2=V\oplus V^\perp$. So, given any $u\in \ell^2$ we have $u=u_1+u_2$ with $u_1\in V$ and $u_2\in V^\perp$. This implies that $$\langle u,w\rangle=\langle u_1,w\rangle,\quad\forall\ u\in \ell ^2.$$

This shows that $$\begin{align} \langle u,w\rangle=0\quad &\Longleftrightarrow\quad\langle u_1,w\rangle=0\\ &\Longleftrightarrow\quad u_1=0\text{ or } w=0\tag{because $u_1,w\in V$}\\ &\Longleftrightarrow\quad u\in V^\perp \tag{because $T\neq 0$} \end{align} $$

So, your question can be rewritten as

Is it true that $u\in \ell ^2\setminus V$ implies $u\in V^\perp$?

As $V\neq\{0\}$ there exists $v\neq 0$ such that $v\in V$. As $V\neq \ell ^2$ there exists $z\neq 0$ such that $z\in V^\perp$. Note that the sum $v+z$ does not belongs to $V$ nor to $V^\perp$. So, the answer is negative (we have $v+w\in\ell^2\setminus V$ and $v+z\notin V^\perp$).

Remark. Exactly the same argument works if we replace $\ell^2$ by any Hilbert space.

$\endgroup$
  • $\begingroup$ thanks @Pedro for yor exhaustive answer $\endgroup$ – Matey Math May 31 '17 at 17:57

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.