# Let $f\in L^{\infty}$, its that true $Pf\in H^{\infty}$?

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 $$L^{\infty}(S^1)$$