Adjoint of multiplication by $z$ in a Hilbert Space (Bergman space) 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?
 A: 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).
