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 $$ \langle u,w\rangle =0 $$


  • $\begingroup$ 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

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 $T(v) = \langle v,e_1\rangle$.

  • $\begingroup$ 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

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.

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

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