Riesz representation theorem in a subspace of $\ell^2$ 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.
 A: 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.
A: 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 $T(v) = \langle v,e_1\rangle$.
