Why is $T_fT_g - T_{fg}$ compact?

Let $ f, g: \mathbb{S}^{1} \to \mathbb{C} $, be continuous functions. Let $ T_{f} $ be the Toeplitz operator with symbol $ f $ acting on the Hardy space $\mathcal{H}$, as defined in this answer.

There it is also written that

If $ f $ and $ g $ are continuous functions on $ \mathbb{S}^{1} $, then it is a basic fact that $ T_{f} T_{g} - T_{fg} $ is a compact operator on $ \mathcal{H} $.

How can we see that this "semi-commutator" of Toeplitz operators is compact?

Also, is this related to the commutator of $T_f$ and $T_g$? Why do we study such object?

Solution 1:

Let $P$ be the projection onto $H^2$.

We have $$ T_fT_g-T_{fg}=PM_fPM_gP-PM_fM_gP=-PM_f(I-P)M_gP. $$ It is enough to show that $PM_f(I-P)$ is compact. We can write $f$ as a norm limit of polynomials. We will see that $PM_p(I-P)$ has finite-rank when $p$ is a polynomial, and then $PM_f(I-P)$ is compact, being a limit of finite-rank operators.

Since a polynomial is a linear combination of nonnegative powers of $z$, it is enough to show that $PM_{z^k}(I-P)$ is finite rank for each $k$. This follows from $$\ PM_{z^k}(I-P)z^m=\begin{cases}z^{k+m},&\ -k\leq m<0\\[0.3cm] 0,&\ \text{otherwise} \end{cases} $$