Note That
$\mathbb{T} = \mathbb{R} / 2 \pi \mathbb{Z}$
$H^2 (\mathbb{T}) = \{ f \in L^2| \hat{f} (n) =0, \forall n <0 \}$.
$P: L^2 (\mathbb{T}) \to H^2(\mathbb{T})$
Prove the following:
1-For $f \in C(\mathbb{T}), [M_f,P] = M_fP - PM_f \in K(L^2 (\mathbb{T}))$.
Hint: First show $[M_f,P] \in F(L^2 (\mathbb{T}))$ for a trigonometric polynomial $f$. Then use the fact that the map $C(\mathbb{T}) \ni f \mapsto [M_f,P] \in B(L^2 (\mathbb{T}))$ is continuous.
2-For $f \in C(\mathbb{T})$ and $g \in L^\infty (\mathbb{T})$, $Tfg - Tf \,Tg \in K(L^2 (\mathbb{T}))$.
Hint: $PMfgP - P Mf P Mg P = PMf(I-P)MgP$.
3-For $f \in C(\mathbb{T})^{-1}$, $Tf \in FR(H^2 (\mathbb{T}))$.
Hint: Use (2) and Atkinson's theorem.
I faced this problem in my functional analysis but I have no idea how start or using the hint. Could you please show some steps about how to tackle this problem.
I assume by $Mf$ you mean the multiplication operator. I will denote it by $M_f$ instead.
As it is known, the trigonometric polynomials are dense in $C(\mathbb{T})$ with the supremum norm. Now let $z:\mathbb{T}\to\mathbb{C}$ be the inclusion function, $z(\zeta)=\zeta$. Suppose that we have shown that, for any $j\in\mathbb{Z}$, the commutator $[M_{z^j},P]$ is a compact operator.
If $f$ is a trigonometric polynomial, then $f=\sum_{j=-N}^Na_jz^j$ and one can easily see that $M_f=\sum_{j=-N}^Na_jM_{z^j}$. Thus $$[M_f,P]=\sum_{j=-N}^Na_j[M_{z^j},P]$$ and this is a compact operator, as a linear combination of compact operators.
Now if $f\in C(\mathbb{T})$ and $\varepsilon>0$, then we can find a trigonometric polynomial $g\in C(\mathbb{T})$ with $\|f-g\|_\infty<\varepsilon$. Then, $$\|[M_f,P]-[M_g,P]\|=\|M_fP-PM_f-M_gP-PM_g\|=\|[M_f-M_g, P]\|\leq$$ $$\leq2\|P\|\|M_f-M_g\|\leq2\varepsilon$$ This shows that we can approximate the operator $[M_f,P]$ with compact operators, so the operator $[M_f,P]$ lies in the closure of the compact operators. But the compacts are a closed set, so $[M_f,P]$ is itself compact.
So all we have to do now is show that $[M_{z^j},P]$ is a compact operator. Note that $L^2(\mathbb{T})$ has the standard ONB $\{\epsilon_j\}_{j\in\mathbb{Z}}$ where $\epsilon_j(\zeta)=\zeta^j$. Now the projection $P$ is defined by $P(\epsilon_j)=0$ if $j\leq0$ and $P(\epsilon_j)=\epsilon_j$ for $j>0$. Also, it is very easy to see that the Multiplication operator $M_{z^k}$ acts as $M_{z^k}(\epsilon_j)=\epsilon_{j+k}$ for all $k,j\in\mathbb{Z}$.
Now we show that $[M_{z^k},P]$ is compact for all $k\in\mathbb{Z}$. Indeed, for all $j\in\mathbb{Z}$ we have that $$[M_{z^k},P](\epsilon_j)=M_{z^k}P(\epsilon_j)-PM_{z^k}(\epsilon_j)=M_{z^k}P(\epsilon_j)-P(\epsilon_{j+k})$$ Let's assume that $k\geq0$.
If $j>0$, then $j+k>0$ and the above becomes $$M_{z^k}P(\epsilon_j)-P(\epsilon_{j+k})=e_{j+k}-e_{j+k}=0.$$ If $j\leq0$ then we distinguish two cases: if $j<-k$, so $j+k<0$, then the above becomes $$M_{z^k}P(\epsilon_j)-P(\epsilon_{j+k})=0-0=0.$$ On the other hand, if $j\in\{-k,-k+1,\dots,-1,0\}$, then $j+k>0$, so $$M_{z^k}P(\epsilon_j)-P(\epsilon_{j+k})=0-\epsilon_{j+k}=-\epsilon_{j+k}.$$ So we see that the only elements of the orthonormal basis $\{\epsilon_j\}$ that can be mapped to non-zero functions are only those $\epsilon_j$ for $j\in\{-k,-k+1,\dots,-1,0\}$ Therefore, the image of the operator $[M_{z^k},P]$ has finite dimensional range, so this operator is finite rank, thus compact.
Similarly one handles the case for $k<0$.
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(2) $T_{fg}-T_fT_g=PM_{fg}P-PM_fPM_gP=PM_fM_gP-PM_fPM_gP=PM_f(I-P)M_gP$. Now since $P$ is a projection, $P(I-P)=(I-P)P=0$, we have that $$PM_f(I-P)M_gP=(PM_f-M_fP)(I-P)(M_gP-PM_g)=-[M_f,P]\cdot(I-P)\cdot[M_g,P]$$ and this is compact because of (1) and the fact that the compact operators are an ideal in $B(H)$.
(3) If $f\in C(\mathbb{T})$ is invertible (which means that $f\neq0$ everywhere, so $g=\frac{1}{f}$ is its well-defined inverse) then by (2) we have that $I-T_fT_g$ and $I-T_gT_f$ are compact operators. By Atkinson's theorem we have that $T_f$ is a Fredholm operator.