My question is: Let $f\in L^{\infty}[S^1]\subseteq L^2[S^1]$, is that always true for $Pf\in H^{\infty}[S^1]\subseteq H^2[S^1]$? Here, $S^1$ is the unit circle in complex plane, and $H^{\infty}$ denote the subspace of $L^{\infty}$ consisting of functions whose $n$-th Fourier coefficients are all zero, $n<0$. Similarly defined $H^2[S^1]$ by the closed subspace of $L^2[S^1]$ consisting of functions whose $n$-th Fourier coefficients are all zero, $n<0$. $P: L^2\mapsto H^2$ denote the projection operator.

Here are my thoughts. This question rises when I study Toeplitz operators on hardy-Hilbert space(GTM 237). It is equivalent to ask: for all $\phi\in L^{\infty}$, is that always true $T_{\phi}1\in H^{\infty}$? I guess the answer is not and try to make a proof by contradict. Let $U$ denote the right shift on $H^2$, if $T_{\phi}1\in H^{\infty}$. Calculation can shows $T_{\phi}U-UT_{\phi}$ has at most rank one, so we can obtain $T_{\phi}e^{in\theta}=T_{\phi}U^n1=U^nT_{\phi}1+F_n1$, where $F_n$ is rank one operator. This shows $T_{\phi}e^{in\theta}$ are all in $H^{\infty}$. But note that $H^{\infty}$ is dense but not closed in $H^2$, so I don’t know how to get a contradiction.

This is in fact a problem in Fourier analysis. So I think it can be solved by use some knowledge in Fourier analysis. Any help or hint? Thanks!


Take $f(\theta)=\sum_{n \ge 1} \frac{\sin n\theta}{n}=\frac{\pi -\theta}{2}, 0 < \theta < 2\pi$; $f \in L^{\infty}(S^1)$ (this could be proved easily summing by parts even if we do not know the result)

However $2iPf=\sum_{n \ge 1} \frac{e^{in\theta}}{n}$ is obviously unbounded near $0$

Actually, even more, is true - if we denote by $\bar H_0^{\infty}(S^1)$ the space of bounded functions with Fourier series with only strictly negative indices coefficients (the conjugate of the functions in $H^{\infty}$ except that we let the constant term to be zero to avoid intersection), $H^{\infty}+\bar H_0^{\infty}$ is far from being dense in $L^{\infty}(S^1)$

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