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I am learning Hilbert space theory from Halmos' "Introduction to Hilbert space and the theory of spectral multiplicity".

While talking about understanding adjoints (p. 39), he calls special attention to this example, remarking that "its adjoint is not what at first it might appear to be":

Let $\mathfrak H$ be the set of analytic functions defined in the interior of the unit disk ($D$), square integrable with respect to planar Lebesgue measure. Then $\mathfrak H$ –called a Bergman space– is a Hilbert space with the inner product $$\langle f,g \rangle= \int_D f\bar g d\lambda = \int_D f(x+iy)\bar g (x+iy) dxdy .$$

In $\mathfrak H$, consider the multiplication by $z$ opertator $A$, i.e. $(Af)(z)=zf(z).$

In the typical $L^2$ I think $A^*$ would simply be multiplication by $\bar z$, but that ruins the differentiability of $f$, so in this case it must be some other thing.

I have thought that this multiplication operator works like the shift operator in sequence spaces if one identifies the function $f$ with its power series terms $(a_0, a_1, ..)$, mapping this sequence to $(0, a_0, a_1,..)$. I know that the usual right shift defined in $l^2$ has as adjoint the left shift when one considers the inner product $\langle \{a_n\},\{b_n\}\rangle = \sum a_n\bar b_n$, but I don't know how the inner product of $\mathfrak H$ would look like translated to the language of its corresponding sequence space, so this approach hasn't helped me much either.

How can I construct this adjoint operator?

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The idea with the sequence space is a good one. A very nice feature of $\mathfrak{H}$ is that its Hilbert space structure fits nicely with Taylor expansion about $0$: The functions $p_n \colon z\mapsto z^n$ are mutually orthogonal.

$$\begin{align} \langle p_k, p_n\rangle &= \int_D z^k\cdot \overline{z^n}\,d\lambda\\ &= \int_0^1 \int_0^{2\pi} r^k e^{ik\varphi}\cdot r^n e^{-in\varphi} \,d\varphi\; r\,dr\\ &= \int_0^1 r^{k+n+1} \left(\int_0^{2\pi} e^{i(k-n)\varphi}\,d\varphi\right)\,dr\\ &= \begin{cases} \dfrac{\pi}{n+1} &, k = n\\\quad\vphantom{\dfrac12} 0 &, k \neq n. \end{cases} \end{align}$$

Since they span a dense subspace, the scaled functions form a Hilbert basis. Let us denote

$$b_n(z) = \sqrt{\frac{n+1}{\pi}}\cdot z^n.$$

Then $(b_n)_{n\in\mathbb{N}}$ is an orthonormal basis of $\mathfrak{H}$, and the multiplication with $z$ in that basis becomes

$$\begin{align} z\cdot f(z) &= z\cdot \sum_{n=0}^\infty c_n\cdot b_n(z)\\ &= z\cdot \sum_{n=0}^\infty c_n\sqrt{\frac{n+1}{\pi}}\cdot z^n\\ &= \sum_{n=0}^\infty c_n\sqrt{\frac{n+1}{\pi}}\cdot z^{n+1}\\ &= \sum_{n=1}^\infty c_{n-1} \sqrt{\frac{n}{\pi}}\cdot z^n\\ &= \sum_{n=1}^\infty c_{n-1}\sqrt{\frac{n}{n+1}}\cdot b_n(z), \end{align}$$

so it's not just a shift of the corresponding sequence, but a shift together with a multiplication of the terms with the sequence $\sqrt{\frac{n}{n+1}}$.

That makes it easy to determine the adjoint in terms of the sequence of coefficients with respect to $(b_n)$. Translating that into the representation by the Taylor series and then to the function-level gives a basis-free characterisation of the adjoint (involving a definite integral).

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  • $\begingroup$ That's quite clarifying! I think then I can see the space $\mathfrak H$ as a sequence space with the product $\langle (a_n),(b_n)\rangle = \sum a_k\bar b_k \frac \pi {k+1}$. In that space the adjoint operator would be defined by $A^*(b_n)=(b_{n+1}\frac{n+1}{n+2})$. I immediately see that the resulting function will be analytic with the same radius of convergence, but, how do I know that it is square integrable? Also, this $A^*$ we found surely is linear, but, is it bounded? $\endgroup$
    – dafinguzman
    Jan 24, 2014 at 17:30
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    $\begingroup$ The square integrability is equivalent to the square summability of the Taylor coefficients with the appropriate weight. That is preserved by the shift, and by multiplication with bounded sequences, so $A^\ast(f)$ is square integrable with $f$. To see the boundedness of $A^\ast$, use Parseval's identity to compute the square of the norm of $A^\ast(f)$ and of $f$. Compare to see that $\lVert A^\ast\rVert \leqslant 1$. $\endgroup$
    – Daniel Fischer
    Jan 24, 2014 at 18:58
  • $\begingroup$ One last detail: I had some confusion relating to convergence issues. You said that your $b_n(z)$ functions were a basis because they "span a dense subspace". $\endgroup$
    – dafinguzman
    Jan 28, 2014 at 23:51
  • $\begingroup$ Yes, although it might be easiest to see that the space of polynomials is dense by verifying that an $f\in\mathfrak{H}$ that is orthogonal to all the $b_n$ must be zero. $$\langle f,b_n\rangle = \lim_{r\nearrow 1} \int_{\lvert z\rvert \leqslant r} f(z)\cdot\overline{b_n(z)}\,d\lambda.$$ On the smaller disk $\lvert z\rvert \leqslant r$, the Taylor series of $f$ converges uniformly, so you can interchange summation and integration, and are left with $\sqrt{\frac{\pi}{n+1}}r^{2n+2}\cdot a_n$ for the integral (if I haven't miscalculated without pen and paper). $\endgroup$
    – Daniel Fischer
    Jan 29, 2014 at 0:15
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    $\begingroup$ $A - \lambda I$ is multiplication by $z-\lambda$. If $\lambda\in\mathbb{D}$, that is not surjective, and if $\lambda\notin\overline{\mathbb{D}}$, it is invertible ($(z-\lambda)^{-1}$ is then bounded on $\mathbb{D}$), so the spectrum is $\overline{\mathbb{D}}$. $\endgroup$
    – Daniel Fischer
    Feb 10, 2014 at 10:46

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