# Find the element representing evaluation at a point in the Hilbert space of $L^2$ holomorphic functions on the disk

The space $S$ of holomorphic functions on the disk $\mathbb{D}$ which are also in $L^2(\mathbb{D})$ is a Hilbert space, with the inner product being $\langle f, g\rangle = \int f(z) \overline{g(z)} dA(z)$, where $dA$ denotes Lebesgue measure. The linear functional $L_{z_0}$, given by $f\mapsto f(z_0)$, is a bounded linear functional on $S$, and thus there must exist some $g_{z_0}\in S$ so that $L_{z_0}(f) = \langle f, g_{z_0}\rangle$ for all $f$. It turns out, this function is given by $$g_{z_0}(w) = \frac{1}{(1-\overline{z_0}z)^2}$$ I have seen this proven by representing $f$ and $g$ as power series, performing the multiplication, and working out the coefficients of the $g$ power series by brute force, which is fine. But I want to think that there is an easier way to do it, or at least a more conceptually satisfying way. For example, I would think to do it by first noting that for $z_0 = 0$, $g_{z_0}$ can be a constant function (I believe $1/\pi$), and then getting the general case by composing with a Mobius transformation; indeed, the function $g_{z_0}$ is closely related to the derivative of such a Mobius transformation, but I have not been able to get such an approach to work. I realized that I don't really know any machinery for working with non-holomorphic functions (like $\overline{g_{z_0}}$), and area integrals of complex functions are also somewhat new to me.

In sum, Can anyone give me a conceptually satisfying way of approaching this problem, hopefully one that uses some classical complex analysis tricks like Cauchy's integral formula, Mobius transformations, and the like?

Or, Can you direct me to a resource which has the basic tools for solving this problem?

Your Möbius transformation idea is good, but it needs one extra thing to work. We know that $$f(0) = \frac{1}{\pi}\iint_{|z|<1} f(z)\,dA(z) \tag1$$ Given $z_0$, apply (1) to $g = (f\circ \phi^{-1})(\phi^{-1})'$ where $\phi(z)= (z-z_0)/(1-\bar z_0z)$ is a Möbius transformation sending $z_0$ to $0$. (Here, $(\phi^{-1})'$ is that extra thing.) Since $$\phi'(z) = \frac{1-\bar z_0 z + \bar z_0(z-z_0) }{(1-\bar z_0 z)^2} = \frac{1- |z_0|^2 }{(1-\bar z_0 z)^2} \tag2$$ the change of variables $z=\phi(w)$ yields $$f(z_0) (\phi^{-1})'(0) = g(0) = \frac{1}{\pi}\iint_{|z|<1} g(z)\,dA(z) = \frac{1}{\pi}\iint_{|w|<1} \frac{f(w)}{\phi'(w)} |\phi'(w)|^2\,dA(w) \tag3$$ This simplifies to $$\frac{f(z_0)}{\phi'(z_0)} = \frac{1}{\pi}\iint_{|w|<1} f(w) \overline{\phi'(w)} \,dA(w) \tag4$$ and, using (2), further simplifies to $$f(z_0) (1-|z_0|^2) = \frac{1}{\pi}\iint_{|w|<1} f(w) \frac{1-|z_0|^2}{(1-z_0 \bar w)^2} \,dA(w) \tag6$$ which is the desired result $$f(z_0) = \frac{1}{\pi}\iint_{|w|<1} f(w) \frac{1}{(1-z_0 \bar w)^2} \,dA(w) \tag7$$

What's the idea of putting the derivative factor $(\phi^{-1})'$ in the formula for $g$? Without it we'd end up with the Jacobian $|\phi'|^2 = \phi'\overline{\phi'}$ as the factor of $f$ on the right hand side of (3). And we want an antiholomorphic weight there, so $\phi'$ has to be canceled off.

• This is beautiful, thank you. That’s exactly what I needed. – Van Latimer Jul 16 '18 at 22:58