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Let $H$ be a Hilbert Space and let $H^*$ be the dual space of $H$.

The Riesz Lemma states that for each $T\in H^*$, there is a unique $y_T\in H$ such that $T(x)=(y_T,x)$ $\forall x\in H$. Also, $||y_T||_H$ = $||T||_{H^*}$.

This doesn't show that there is a one-to-one correspondence between the hilbert space $H$ and its dual space $H^*$.

So, to derive the one-to-one correspondence between $H$ and $H^*$, we need to show that for each $y\in H$, then there exists an unique $T_y \in H$ such that $T_y(x)=(y,x)$ $\forall x \in H$.

I was thinking about showing that $T_y(x)=(y,x)$ for some subspace(s) (whose span is $H$). But I wasn't successful in doing this. I'm also thinking about the Cauchy-Schwarz Inequality to show an equality (by showing an inequality in both directions...)

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I'm not sure I understand the question. The map $H \to H^{\ast}$ given by $y \mapsto (y,\cdot)$ is clearly injective (it is isometric, for example) and it is surjective by what you call the Riesz lemma. –  t.b. Jul 25 '11 at 7:11
    
By the way: By stating "For each ... there exists a unique ..." you are already implying that there is a bijection. To spell it out completely: "for each ... there exists ..." implies onto, "unique" implies one-to-one. –  t.b. Jul 25 '11 at 7:20
    
Thanks Theo. Do you then think there is any point in proving the converse I stated above: for each y∈H, then there exists an unique Ty∈H such that Ty(x)=(y,x) ∀x∈H. ? –  r.g. Jul 25 '11 at 7:52
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Well, there is nothing to prove $T_y(\cdot) = (y,\cdot)$. Every other linear functional is of the form $(y', \cdot)$ by the Riesz lemma. Since for $z = y - y'$ we have $(y,z) - (y',z) = \|z\|^2$ we must have $z = 0$ if the two functionals are the same, i.e. $y = y'$. –  t.b. Jul 25 '11 at 8:08
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"for each y∈H, then there exists an unique Ty∈H such that Ty(x)=(y,x) ∀x∈H": Well, $T_y$ is defined by the above formula; you just have to check that it's a linear and bounded functional (which is quite easy). It's like asking "Why is there a (unique) function $f$ such that $f(x)=x^2 \forall x\in\mathbb{R}$" –  Florian Jul 25 '11 at 8:34

1 Answer 1

up vote 4 down vote accepted

I still don't understand the source of confusion here. You can find a proof of the Riesz representation theorem in (the second part of) my answer here which is essentially the same as the proof of Theorem 3.2.3 on p.89 of Pedersen's Analysis Now. Any other basic book on functional analysis will contain a proof.

Look: The Riesz lemma you state tells you that every linear functional $T \in H^{\ast}$ is of the form $T(x) = (y_T,x)$ for a unique $y_T \in H$.

This means that the map $H \to H^{\ast}$ given by $y \mapsto (y,\cdot)$ is

  1. onto — because every continuous linear functional is of this form, and it is
  2. one-to-one — either because it is isometric (as $\|T\| = \|y_T\|$) — or directly because for $y \neq y'$ we can take $z = y - y' \neq 0$, hence $(y,z) - (y',z) = (z,z) = \|z\|^2 \neq 0$, so $(y,\cdot) \neq (y',\cdot)$.

Now the lemma you want to prove is a triviality: we already have a bijection $y \to (y,\cdot)$ between $H$ and $H^{\ast}$.

Given $y \in H$ the map $T_y$ defined by $x \mapsto T_y(x) = (y,x)$ is a linear functional and it is already of the desired form $T_y = (y,\cdot)$. It is continuous of norm $\|T_y\| \leq \|y\|$ by Cauchy–Schwarz and plugging $y$ into the definition gives $T_y(y) = (y,y) = \|y\|^2$, so we see that $\|T_y\| \geq \|y\|$, hence $\|T_y\| = \|y\|$.

So $T_y = (y,\cdot)$ is a continuous linear functional and by repeating the second argument in 2. above we see that we must have $y_{T_y} = y$ (we can appeal to the uniqueness part of the Riesz lemma, too of course).

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