M. Riesz theorem on conjugate functions

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The conformal mapping of UU onto a vertical strip shows that M. Riesz's theorem on conjugate functions cannot be extended to p=∞p = \infty. Deduce that it cannot be extended to p=1p = 1 either.

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Problem 2.17.25 from Rudin W. "Real and Complex Analysis" [Rud87]More problems in Chapter 2.17 "$H^p$ Spaces"

Suppose 1<p<∞1 < p < \infty, and associate with each f∈Lp(T)f \in L^{p}(T) its Fourier coefficients

f^(n)=12π∫−ππf(eit)e−intdt(n=0,±1,±2,…) \hat{f}(n) = \frac{1}{2 \pi } \int _{-\pi }^{\pi } f\left(e^{i t}\right) e^{-i n t} d t \quad (n = 0, \pm 1, \pm 2, \ldots )

Deduce the following statements from Theorem 17.26:

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(a)

To each f∈Lp(T)f \in L^{p}(T) there corresponds a function g∈Lp(T)g \in L^{p}(T) such that g^(n)=f^(n)\hat{g}(n) = \hat{f}(n) for n≥0n \geq 0 but g^(n)=0\hat{g}(n) = 0 for all n<0n < 0. In fact, there is a constant CC, depending only on pp, such that

∥g∥p≤C∥f∥p \left\| g\right\| _{p} \leq C\left\| f\right\| _{p}

The mapping f→gf \rightarrow g is thus a bounded linear projection of Lp(T)L^{p}(T) into Lp(T)L^{p}(T). The Fourier series of gg is obtained from that of ff by deleting the terms with n<0n < 0.

(b)

Show that the same is true if we delete the terms with n<kn < k, where kk is any given integer.

(c)

Deduce from (b) that the partial sums sns_{n} of the Fourier series of any f∈Lp(T)f \in L^{p}(T) form a bounded sequence in Lp(T)L^{p}(T). Conclude further that we actually have

lim⁡n→∞∥f−sn∥p=0 \lim _{n \rightarrow \infty }\left\| f-s_{n}\right\| _{p} = 0
(d)

If f∈Lp(T)f \in L^{p}(T) and if

F(z)=∑n=0∞f^(n)zn F(z) = \sum _{n = 0}^{\infty } \hat{f}(n) z^{n}

then F∈HpF \in H^{p}, and every F∈HpF \in H^{p} is so obtained. Thus the projection mentioned in (a) may be regarded as a mapping of Lp(T)L^{p}(T) onto HpH^{p}.

Show that there is a much simpler proof of Theorem 17.26 if p=2p = 2, and find the best value of A2A_{2}.

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References

[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.