$\newcommand{\A}{A({\mathbb D})}\newcommand{\C}{C({\mathbb T})}$Let $\C$ the Banach space consisting of all complex
valued, continuous functions on the unit circle $\mathbb T$ under the sup norm. The disk algebra $\A$ is an intensely
studied subspace of $\C$ which can be defined in three equivalent ways:
-
The closed subspace of $\C$ spanned by the functions $z\mapsto z^n$, for $n\geq 0$.
-
The subset of $\C$ formed by all functions $f$ with vanishing negative
Fourier coefficients, that is,
$$
\hat f(n)=0,\quad\forall n<0.
$$ -
The subset of $\C$ formed by all functions $f$ admitting a continuous extension $\tilde f$ to the closed unit disk $\mathbb D$,
such that $\tilde f$ is
holomorphic on the interior of $\mathbb D$.
It is a fact that $\A$ is not complemented in $\C$ in the sense that there is no closed subspace $E\subseteq \C$ such that
$$
\C=\A\oplus E.
$$
This can be shown to be equivalent to the following statement: There exists a doubly infinite sequence of complex numbers $\{a_n\}_{n=-\infty }^\infty $, such that
$$
\sum_{n=-\infty }^\infty a_n e^{int}
$$
is the Fourier series of a continuous, $2 \pi $-periodic function, while the positive part of this series, namely
$$
\sum_{n=0}^\infty a_n e^{int}
$$
is not.
Question. What is a concete example of a sequence $\{a_n\}_{n=-\infty }^\infty $, as above?
NOTES
See On projection of fourier series for the corresponding result
regarding $L^\infty(\mathbb T)$.
I found two proofs of the fact that $\A$ is not complemented in $\C$, which would in theory be enough to provide an
answer to my question, but I was unable to make it work. These proofs can be found in:
-
Page 137 of Rudin, Walter, Functional analysis, McGraw-Hill Series in Higher Mathematics. New York etc.: McGraw-Hill Book Comp. XIII, 397 p. (1973). ZBL0253.46001.
-
Page 55 of Hoffman, K., Banach spaces of analytic functions, Prentice-Hall Series in Modern Analysis. Englewood Cliffs, N.J.: Prentice-Hall, Inc. XIII, 217 p. (1962). ZBL0117.34001.
Best Answer
Pick $b_n$ decreasing to zero, $nb_n \to 0, \sum b_n =\infty$; for example $b_n=\frac{1}{n\log n}, n \ge 2$ will do.
Then it is not hard to show that $g(x)=\sum_{n \ge 2} b_n\sin nx$ is continuous on the reals since the series converges uniformly everywhere.
(the result is true but requires a summation by part and a classical estimate on the partial sums of $\sum \sin nx$ in general when $nb_n \to 0$ but not necessarily monotonically, however for the specific case above where $nb_n$ decreases to zero, it immediately follows from the Abel-Dirichlet convergence criterion for series since it is well known that $\sum_{n \ge 2} \frac{\sin nx}{n}$ has partial sums uniformly bounded in $x$)
However $h(x)=\sum_{n \ge 2} b_n \cos nx$ is not continuous since the series is divergent to $\infty$ at $0$ and $h$ is continuous on $(0, 2\pi)$ but unbounded (though integrable of course) at $0$
But then taking $f(x)=\sum_{n \in \mathbb Z, |n| \ge 2}(-i(\text {sign n}) {b_{|n|}})e^{inx}=2g(x)$ is continuous, but its projection is obviously not since its imaginary part is not by the above