Toeplitz operator on Bergman space

Question

Let $A^2(\mathbb{D})$ be the Bergman space over the open unit disk in $\mathbb{C}$, i.e. the set of holomorphic square integrable functions on $\mathbb{D}$. Denote by $P^{A^2}$ the orthogonal projection from $L^2(\mathbb{D})$ onto $A^2(\mathbb{D})$, and for $\phi \in L^2(\mathbb{D})$, we define the Toeplitz operator $T_{\phi}: A^2(\mathbb{D}) \rightarrow A^{2}(\mathbb{D})$ by: \begin{equation} T_{\phi}: f \mapsto P^{A^2}(\phi f). \end{equation} Now define $\phi: \mathbb{D} \rightarrow \mathbb{C}$ by $\phi(re^{i\theta}) = \frac{1}{\sqrt{r}} \chi_{(0, \frac{1}{2}]}(r)$. I am looking to show that $T_\phi$ is a bounded operator on $A^2(\mathbb{D})$, but I can't go further than writing out the norm in $A^2(\mathbb{D})$. Is there another (perhaps a more "explicit") way to define the Toeplitz operator on $A^2(\mathbb{D})$?

Answer 1

Hint: You know the Bergman kernel, so you know we need to show the function of $z$ given by

$$\tag 1 \int_{\mathbb D} \frac{\phi (w)f(w)}{(1-z\bar w)^2}\,dm_2(w)$$

is in $A^2,$ with appropriate bounds. Here $m_2$ is Lebesgue measure in the plane. Going to polar coordinates, we see $(1)$ equals

$$\tag 2 \int_0^{2\pi}\int_0^{1/2} \frac{f(re^{it})}{r^{1/2}(1-zre^{-it})^2}\,r\,dr\, dt.$$

There is lots of room here, so crash through with absolute values at will and see what happens.

Answer 2

I have found an alternative way to solve the problem, so I'll add it here for the sake of completeness. I first remark that $\phi f \in L^2(\mathbb{D})$ for all $f \in A^2(\mathbb{D})$ - this is easily proven by writing out the $L^2(\mathbb{D})$-norm and switching to polar coordinates. Thus, for all $f \in A^2(\mathbb{D})$ we have $$P^{A^2}(\phi f) \in A^2(\mathbb{D})$$ since the orthogonal projection is well defined. Now we recall that $A^2(\mathbb{D})$ is a separable Hilbert space, and its orthonormal basis is given by the vectors $(e_n)_{n \in \mathbb{N}}$ defined, for all $n \in \mathbb{N}$ by \begin{equation} e_n: z \mapsto \sqrt{n+1}z^n \end{equation} (I have chosen to use the normalized Lebesgue measure over the disk to avoid a $\sqrt{\pi}^{-1}$ factor in the basis). This means that any $f \in A^2(\mathbb{D})$ can be written in the form \begin{equation} f = \sum_{n = 0}^{\infty} \langle f, e_n \rangle e_n, \end{equation} the limit being taken with respect to the $A^2(\mathbb{D})$-norm. In particular, \begin{equation} T_{\phi} f = P^{A^2}(\phi f) = \sum_{n=0}^{\infty} \langle P^{A^2}(\phi f), e_n \rangle e_n = \sum_{n=0}^{\infty} \langle \phi f, P^{A^2} e_n \rangle e_n = \sum_{n=0}^{\infty} \langle \phi f, e_n \rangle e_n \end{equation} (here I used the self-adointness of $P^{A^2}$ and the fact that $e_n \in A^{2}(\mathbb{D})$ for all $n \in \mathbb{N}$). This form allows us to prove the boundedness and compactness of $T_\phi$ in a more "direct" way. To prove that $T_\phi$ is bounded, take an arbitrary $f \in A^2(\mathbb{D})$. By Parseval's theorem, \begin{equation} \|T_{\phi} f\|_{A^2}^2 = \sum_{n=0}^{\infty} |\langle \phi f, e_n \rangle|^2. \end{equation} We now look at $|\langle \phi f, e_n \rangle|^2$. The Cauchy-Schwarz inequality gives \begin{equation} |\langle \phi f, e_n \rangle|^2 \leq \|f \|_{A^2}^2 \|\phi e_n \|_{A^2}^2, \end{equation} so the only thing that remains to be shown is that the (numerical) series $\sum_{n=0}^{\infty} \|\phi e_n \|_{A^2}^2$ is convergent. We calculate \begin{equation} \int_{\mathbb{D}} |\phi(z)e_n(z)|^2 dm(z) = (n+1) \int_{0}^{2\pi} \int_{0}^{\frac{1}{2}} r^{-1} r^{2n} |e^{int}|^2 r \ dr \ dt = \frac{\pi(n+1)}{2n+1} \Big(\frac{1}{2}\Big)^{2n}, \end{equation} so we obtain in fact a convergent geometric series. Since $f$ was taken arbitrary, we conclude that \begin{equation} \|T_{\phi} \| \leq \Big(\pi \sum_{n=0}^{\infty} \frac{n+1}{2n+1} \frac{1}{4^n} \Big)^{\frac{1}{2}} < +\infty. \end{equation} We now prove that $T_{\phi}$ is in fact compact. In order to do this, we will prove that $T_{\phi}$ is the limit of a sequence of finite rank operators. Fix $N \in \mathbb{N}^*$, and define $T_N: f \mapsto \sum_{n=0}^{N} \langle \phi f, e_n \rangle e_n$ from $A^2(\mathbb{D})$ to $A^2(\mathbb{D})$. Observe that every $x \in T_N(A^2(\mathbb{D}))$ is of the form \begin{equation} x = \sum_{n=0}^{N} a_n e_n, \end{equation} where $a_n = \langle \phi f, e_n \rangle \in \mathbb{C}$. Since $(e_n)_{n \in \mathbb{N}}$ form a basis, they are linearly independent. So $\dim(T_N(A^2(\mathbb{D}))) \leq N+1$, and consequently $T_N$ are of finite rank. To prove that $\|T_N - T_{\phi}\| \rightarrow 0$ as $N \rightarrow \infty$, we observe that for an arbitrary $f$ in the unit ball of $A^2(\mathbb{D})$, \begin{equation} \|T_N f - T_{\phi} f \|_{A^2}^2 = \sum_{n > N} |\langle \phi f, e_n \rangle |^2 \leq \sum_{n > N} \|\phi e_n \|_{A^2}^2, \end{equation} and the RHS goes to $0$ as $N \rightarrow \infty$ as the remainder of a convergent geometric series.