A research problem on which I am currently working requires a construction in topological dynamics of the following type:



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> Let $T \colon X \to X$ be a continuous transformation of a compact metric space which contains at least two points, and let $(a\_n)$ be an absolutely summable real sequence which is not the zero sequence. When can we guarantee that there exists a continuous function $f \colon X \to \mathbb{C}$ such that $\sum\_{n=1}^\infty a\_n f \circ T^n$ is not a constant function?
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The above question would seem to amount to a question about the spectral behaviour of the composition operator $U\_T \colon C(X) \to C(X)$ defined by $U\_Tf(x):=f(T(x))$. It is easy to show that the eigenvalues of $U\_T$ form a subgroup of the unit circle. If $U\_T$ has an eigenvalue which is not a root of unity then the eigenvalues are dense in the unit circle, so given a fixed sequence $(a\_n)$ an easy Fourier analysis argument allows us to choose an eigenfunction $f$ for which $\sum\_{n=1}^\infty a\_n f \circ T^n$ is not constant. On the other hand, in some cases where $U\_T$ has a root of unity as an eigenvalue, there are sequences such that the series converges to a constant for all $f$: for example, if $X$ contains just two points then the sequence given by $a\_1=a\_2=1$ and $a\_n=0$ for $n \geq 3$ has this property.
The case of most interest, then, is that in which $U\_T$ has no eigenvalues except $1$ and no eigenfunctions other than the constant function, which is referred to as *topological weak mixing*. Specifically, I ask:



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> Is topological weak mixing of $T$ sufficient to guarantee the existence of $f$ in the first question?
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This suggests to me the following more general functional-analytic question, which (by considering the action of $U\_T$ on the quotient of $C(X)$ modulo the subspace of constant functions) would be sufficient for a positive answer to the above:



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> Let $L$ be a bounded linear operator acting on an infinite-dimensional Banach space $B$, with the spectrum of $L$ being $\{1\}$ and the norm of $L$ being $1$. When does there exist $x \in B$ such that the sequence $\{L^nx \colon n \geq 0\}$ is $\omega$-linearly independent, i.e. for all nonzero absolutely summable sequences $(a\_n)$, the sum $\sum\_{n=1}^\infty a\_n L^nx$ is nonzero?
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Thanks in advance!



Topological weak mixing and $\omega$-linearly-independent sequences generated by composition operators
For each $k > 0$, $F\_k = \{ x \in X \mid T^k(x) = x \}$ is closed, hence if $T$ is not periodic then for each $N > 0$, $G\_N = X \setminus (\cup\_{k \leq N} F\_k)$ is open, and non-empty.


If we are given $(a\_n)$ absolutely summable, (for which we will assume $a\_0 = 1$ by scaling and shifting, which is fine since $T$ is not periodic), take $N > 0$ such that $\sum\_{n > N} |a\_n| < 1$. Since $G\_N$ is open and non-empty there exists a non-empty, open subset $O \subset G\_N$ such that $T^n(O) \cap O = \emptyset$ for all $1 \leq n \leq N$.


If we consider a continuous function $f$ such that $\| f \|\_\infty \leq 1$, $f$ has support contained in $O$, and $f$ obtains the values $\pm 1$ within $O$, then $\sum\_{n = 0}^N a\_nf \circ T^n$ agrees with $f$ on $O$. Hence $\sum\_{n = 0}^\infty a\_n f \circ T^n$ has both positive and negative values and is therefore not constant.


