2
$\begingroup$

Let $\cal H$ be a Hilbert space, and let $\cal H^\ast$ be its dual (of the continuous functionals).

If $\cal H$ is a real vector space, I can define: $$\begin{align}\Phi\colon\, &{\cal H} \to {\cal H}^\ast \\ &x \mapsto \Phi(x)\colon {\cal H} \to \Bbb R \\ &\qquad\qquad\quad y\mapsto \Phi(x)(y) = \langle x,y \rangle\end{align}$$Since $\langle \cdot, \cdot \rangle$ is bilinear, both $\Phi$ and $\Phi(x)$ are linear. Also: $$|\Phi(x)(y)| = |\langle x,y\rangle| \leq \|x\|\|y\|$$gives that $\Phi(x)$ is bounded with $\|\Phi(x)\| \leq \|x\|$. If $x = 0$ there's nothing to do. Otherwise evaluating at $y = x/\|x\|$ gives $\|\Phi(x)\| = \|x\|$. So $\Phi$ preserves norms and so is injective. The Riesz representation theorem gives surjectivity. All good, ${\cal H}\cong {\cal H}^\ast$.

My problem is copying that idea for the case that ${\cal H}$ is a complex vector space. We'll have that $\Phi(x)$ will be linear-conjugate and $\Phi$ will be sesquilinear instead of bilinear. I thought of using $\langle x,-iy\rangle$ instead.. this fixes the conjugacy of the scalars but messes up linearity.

I don't know how to fix this. Can someone help? Thanks.

$\endgroup$
  • $\begingroup$ You want a linear isometry between the Hilbert space and its dual in the complex case? $\endgroup$ – Daniel Fischer Jun 21 '15 at 22:58
  • $\begingroup$ @Daniel sorry the delay in answering, I went out to eat and it's almost 9 PM here. Yes, that's what I want, but all I was getting was a linear-conjugate one. Now let me read your answer :) $\endgroup$ – Ivo Terek Jun 21 '15 at 23:46
6
$\begingroup$

The Riesz map $\Phi\colon y \mapsto \langle\,\cdot\,, y\rangle$ is a conjugate-linear isometric bijection between a complex Hilbert space and its dual. This map is natural, it can be defined without choosing a basis. If $F\colon \mathcal{H}\to \mathcal{H}^\ast$ is a complex-linear isometric isomorphism, then $\sigma \colon \mathcal{H} \to \mathcal{H}$ given by $\sigma(y) = \Phi^{-1}(F(y))$ is a conjugate-linear isometry of $\mathcal{H}$. Conversely, every conjugate-linear isometry $\sigma$ of $\mathcal{H}$ induces a complex-linear isometric isomorphism by composing it with $\Phi$, namely $y \mapsto \langle\,\cdot\,, \sigma(y)\rangle$.

Of particular interest are the cases where $\sigma$ is an involution (i.e. $\sigma \circ \sigma = \operatorname{id}_{\mathcal{H}}$). Then we call $\sigma$ a conjugation on $\mathcal{H}$. Since every Hilbert space is isometrically isomorphic to $\ell^2(\mathcal{B})$ when $\mathcal{B}$ is an orthonormal basis of $\mathcal{H}$, on every complex Hilbert space there exist conjugations, we can pull back the natural conjugation from $\ell^2(\mathcal{B})$, so for every complex Hilbert space there are complex-linear isometric isomorphisms between the space and its dual.

In general, such a conjugation cannot be defined without recourse to a basis, so there is no "natural" conjugation on $\mathcal{H}$, and no "natural" complex-linear isometric isomorphism to the dual.

However, in many cases the Hilbert spaces one considers are spaces of complex-valued functions where the natural conjugation of functions is an isometry, and in those cases we have a "natural" conjugation on $\mathcal{H}$, and a "natural" complex-linear isometric isomorphism $\mathcal{H}\to \mathcal{H}^\ast$ ("natural" means we can define it without recourse to a basis). In $L^2(\mu)$ for a positive measure $\mu$ - that subsumes the $\ell^2(S)$ spaces - the bilinear pairing

$$(f,g) \mapsto \int f\cdot g\,d\mu$$

induces a natural linear isometry between $L^2(\mu)$ and $L^2(\mu)^\ast$. In the Sobolev spaces $H^m(\Omega)$, the bilinear pairing

$$(f,g) \mapsto \sum_{\lvert\alpha\rvert_1 \leqslant m} \int_\Omega D^\alpha f(x) D^\alpha g(x)\,d\lambda(x)$$

induces a natural complex-linear isometry between $H^m(\Omega)$ and $H^m(\Omega)^\ast$, since partial differentiation is a real operator, i.e. $D^\alpha \overline{g} = \overline{D^\alpha g}$.

However, on the Hilbert space

$$H^2(\mathbb{D}) = \biggl\{ f\in \mathscr{O}(\mathbb{D}) : \int_\mathbb{D} \lvert f(z)\rvert^2 \,d\lambda < +\infty\biggr\}$$

of square-integrable holomorphic functions on the unit disk, the natural conjugation of functions leads out of the space - $f$ and $\overline{f}$ are both holomorphic if and only if $f$ is (locally) constant - and thus doesn't induce a natural conjugation on $H^2(\mathbb{D})$. But there still is a natural conjugation on that space, it is given by $\sigma(f) \colon z \mapsto \overline{f(\overline{z})}$, and so we still have a natural complex linear isometric isomorphism between $H^2(\mathbb{D})$ and $H^2(\mathbb{D})^\ast$.

The fact that gives us the natural conjugation is that the domain is invariant under complex conjugation, $z \in \mathbb{D}\iff \overline{z}\in \mathbb{D}$. If we consider a bounded domain $U\subset \mathbb{C}$ that is not invariant under complex conjugation, the above construction no longer works, and if one takes a domain $U$ that is not conformally equivalent to a symmetric (with respect to the real axis) domain, I don't know of a way to define a conjugation on $H^2(U)$ - and thus a complex-linear isometric isomorphism between $H^2(U)$ and $H^2(U)^\ast$ - without recourse to a (Hilbert) basis.

$\endgroup$
  • $\begingroup$ Thanks for the answer. Interesting, I didn't knew about the result for $\ell^2({\cal B})$. But I'm not sure, though, that it addresses the issue at hand: is it possible to have a linear isometry instead of a linear-conjugate one? $\endgroup$ – Ivo Terek Jun 21 '15 at 23:50
  • $\begingroup$ It seems I wasn't expressing myself as clearly as I wanted [well, it was past 1 a.m. for me, that might count as a mitigating circumstance]. Yes, there always exists a complex-linear isometric isomorphism, but in general, such an isomorphism is not natural; like isomorphisms between a finite-dimensional space and its dual are not natural, you need to choose a basis, whereas the isomorphism to the bidual is then natural. I've added some examples to the answer too. $\endgroup$ – Daniel Fischer Jun 22 '15 at 9:16
  • $\begingroup$ I guess I'm hopeless, then. You made your point, thanks! $\endgroup$ – Ivo Terek Jun 22 '15 at 9:54

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.