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In my notes, I have this version of the Riesz representation theorem: Let $X$ be a separable Hilbert space, and let $\{\phi_n\}_n\subset X$ a countable orthonormal set. Let $\{c_k\}_k\in l^2$ a sequence. There exist an element $x\in X$ such that $\forall k\in\mathbb{N},\ (x\mid\phi_k)=c_k$.

My question is: with these hypothesis, is there a unique $x\in X?$. I think that with the additional hypothesis that $\{\phi_n\}_n$ is complete, then I have the uniqueness: as a matter of fact if there were two elements $x,y\in X$ that verify the theorem, then I would have $$(x-y\mid \phi_k)=0\ \forall k,$$ Which would imply $$x-y\in clos(\{Span(\{\phi_k\}_k\})^\perp=X^\perp,$$ and then $$(x-y\mid x-y)=0\Rightarrow x-y=0.$$ Is this right?

2 Answers2

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That's correct. On the other hand, $x$ is not unique if $\{\phi_n\}$ is not complete. Indeed, if $v\in X$ is any nonzero element that is orthogonal to every $\phi_n$, then $y=x+v$ will also satisfy $(y\mid\phi_k)=c_k$ for all $k$.

Eric Wofsey
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You can also look at it in terms of the classic version of the Riesz Representation Theorem.

If $(e_n)_{n=1}^\infty$ is a complete sequence (i.e. an orthonormal basis for $H$), then for any sequence $(c_n)_{n=1}^\infty \in \ell^2$ you can define a bounded linear functional $\phi$ as $e_n \mapsto c_n$ and extend it by linearity and continuity.

Now, according to the classic Riesz Representation Theorem, there exists a unique $v \in H$ such that $\phi(x) = \langle x, v\rangle$ for all $x \in H$.

In particular $c_n = \phi(e_n) = \langle e_n, v \rangle$ for all $n \in \mathbb{N}$.

On the other hand, if $(e_n)_{n=1}^\infty$ is not complete sequence, then $e_n \mapsto c_n$ does not determine a bounded linear functional uniquely, as $\overline{\operatorname{span}}\{e_n\}_{n\in\mathbb{N}} \ne H$.

Therefore $(e_n)_{n=1}^\infty$ is in fact contained in an orthonormal basis $(f_n)_{n=1}^\infty$. Now just define $\phi$ as $e_n \mapsto c_n$ and $f_n \mapsto $ whatever, for any $f_n \ne e_k, \forall k\in\mathbb{N}$, just taking care that $(\phi(f_n))_{n=1}^\infty$ is a sequence in $\ell^2$.

Any such funtional will again satisfy $c_n = \phi(e_n) = \langle e_n, v \rangle, \forall n \in \mathbb{N}$, with a different $v \in H$ for each one.

mechanodroid
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