0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$
  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 $f$".
$\endgroup$
  • $\begingroup$ I thought it was $f \in H$? $\endgroup$ – Al jabra Apr 10 '16 at 17:19
  • $\begingroup$ the problem is I can not see the connection between $H$ and $H^*$ in the theorem. $\endgroup$ – Al jabra Apr 10 '16 at 17:21
  • $\begingroup$ @Aljabra No, the function $f$ is defined on $H$, so it is in $H^*$. Did that answer your question or would you like more elaboration? $\endgroup$ – Ian Apr 10 '16 at 17:23
  • $\begingroup$ I see now, that is from the definition of a dual space. From wikipedia the dual space $H^*$ is the set of all linear functional on $H$. From the Kreyszig book "introductory functional analysis with applications", the dual space $H^*$ is defined as the set of all bounded linear functional on $H$. Which definition is right? And we dont have a bounded linear functional do we? $\endgroup$ – Al jabra Apr 10 '16 at 17:28
  • $\begingroup$ @Aljabra There are distinct notions of "algebraic dual space" and "continuous dual space". The former consists of all linear functionals while the latter consists of just the bounded linear functionals. The one of interest in Hilbert space theory is the continuous dual space. Cauchy-Schwarz tells you that the functional $f(u)=\langle u,v \rangle$ is bounded. (As it happens, any linear functional on a complete normed linear space that you can explicitly write down is bounded, though one can "construct" unbounded linear functionals on complete normed linear spaces using the axiom of choice.) $\endgroup$ – Ian Apr 10 '16 at 17:33

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.