# Hilbert space isometric to a subspace of its dual

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.

• You want a linear isometry between the Hilbert space and its dual in the complex case? – Daniel Fischer Jun 21 '15 at 22:58
• @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 :) – Ivo Terek Jun 21 '15 at 23:46

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.

• 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? – Ivo Terek Jun 21 '15 at 23:50
• 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. – Daniel Fischer Jun 22 '15 at 9:16
• I guess I'm hopeless, then. You made your point, thanks! – Ivo Terek Jun 22 '15 at 9:54