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I have seen in class that for any homogeneous Banach space B $S_N[f] \rightarrow f$ in norm iff B admits conjugaison, I.e. $\forall f \in B,$ exists a $f^* \in B$ with $S[f^*]=S^*[f]=\sum_{n \in \mathbb{z}} -i sgn(n) \hat{f}(n) e^{int}$

Note that I am working here on the circle $\mathbb{T}$.

I have also seen that $L^{\infty}(\mathbb{T})$ and $ C(\mathbb{T}) $ are examples where $S_N$ do not converge to f in norm, hence they must not admit conjugaison.

Can anyone give me two examples:

1) function which does not admit conjugaison on $C(\mathbb{T})$

2) another function which does not admit conjugaison on $L^{\infty}(\mathbb{T})$

usere5225321
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1 Answers1

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Extend the function $f$ on $\mathbb{T}$ to a harmonic functions $u$ on the unit disk $\mathbb{D}$. Then conjugation means looking at the conjugate harmonic function $v$, which is defined (up to an additive constant) by the requirement that $u+iv$ is holomorphic in $\mathbb{D}$. Now we are looking for:

  1. A holomorphic function on $\mathbb{D}$ whose real part has continuous extension to $\mathbb{T}$ but the imaginary part does not.

  2. A holomorphic function on $\mathbb{D}$ whose real part is bounded but the imaginary part is not.

The second one is easier: $i\log(1-z)$ does the job, because $$i\log(1-z) = -\arg(1-z) + i \log|1-z|$$

For the first one, an example is given by a conformal map of $\mathbb{D}$ onto the domain between the imaginary axis and a curve that's asymptotic to it in both directions, e.g., $x=1/(1+y^2)$. The boundary values of the conformal map trace the boundary counterclockwise. The real part varies continuously from 1 to 0 and back. The imaginary part is discontinuous and even unbounded; thus, this example would also work for question 2.

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