[Math] Identifying the dual space of a Hilbert space using the Riesz representation theorem.

functional-analysisriesz-representation-theorem

I know the Riesz representation theorem for a Hilbert space over $\mathbb{C}$:

Let $H$ be a Hilbert space. Then every continuous linear functional $f$ on $H$ is of the form $$f(u) =\langle u,v \rangle $$ where $u \in H$ and where $v$ is a uniquely determined element $v=v_f \in H$.

I am trying to understand how we can identify the dual space $H^*$ of a Hilbert space $H$ using the Riesz representation theorem.

This is the proof.

I am told that from the theorem we can identify each $f \in H^*$ with a unique $v_f \in H$ for which $f =\langle \cdot , v_f \rangle$

Conversely each $v \in H$ induces a continuous linear functional $$f_v : u \to \langle u, v \rangle$$ where $u \in H$.

This means that there is a one to one between elements of $H$ and and those of $H^*$ so we can identify $H^*$ with $H$.

My first question: In the theorem where is the dual space $H^*$
mentioned? I am not sure how we are told from the theorem that we can identify each $F \in H^*$.

My second question: What is the proof trying to say, what is it's general direction?

My third question: What does $f =\langle \cdot , v_f \rangle$ mean?

The problem is I can not see the connection betwen $H$ and $H^*$ in the theorem.

Best Answer

  1. You can restate the theorem as "for every $f \in H^*$ there exists a unique $v_f \in H$ such that $f(u)=\langle u,v_f \rangle$ for all $u \in H$."
  2. Which proof? Are you asking about the proof of "we can identify $H^*$ with $H$" from the statement I just gave? The point is that each element of $H^*$ corresponds to an element of $H$ (by the statement above) while each element of $H$ corresponds to an element of $H^*$ (by bilinearity of the inner product and Cauchy-Schwarz).
  3. $f=\langle \cdot,v_f \rangle$ means "$f(u)=\langle u,v_f \rangle$ for all $u$ in the domain of $f$".
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