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

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?

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:

Is topological weak mixing of $T$ sufficient to guarantee the existence of $f$ in the first question?

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:

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?

Thanks in advance!

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I don't think your last question states what you want--the identity operator is a counterexample. – Bill Johnson Nov 8 '11 at 17:33
This also suggests that under your definition, the identity map $T=\iota$ is topological weak mixing. I thought that latter property was usually defined as $f\in C(X), f\circ T=\lambda f \implies f\in\mathbb C 1$ (I think?) That is, $T$ is topological weak mixing if the only eigenvectors of $U_T$ are the constants. If $U_T$ acts on $C(X)$ quotiented by the constants, then do you want that $U_T$ has no eigenvectors? – Matthew Daws Nov 8 '11 at 18:20
Thanks Bill, Matthew: I've edited the question to address these mistakes. – Ian Morris Nov 8 '11 at 22:05
up vote 3 down vote accepted

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.

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Thanks, that's a great answer. In some cases it is a little nontrivial to choose the open set $O$ in such a way that it contains two distinct points: for example, if $X$ is the one-point compactification of $\mathbb{Z}$, $T(n):=n+1$ for $n \in \mathbb{Z}$ and $T$ fixes the point at infinity, then $O$ cannot simply by a ball around a point $x$. However I think it is sufficient to let $O$ be the set of all points sufficiently near to a set of the form $\{x\}\cup \{T^{N+1}x\}$. – Ian Morris Nov 9 '11 at 12:34
That's right $O$ should have two points. If $G_N$ is discrete then you can even take equality $O = \{ x \} \cup \{ T^{N + 1}(x) \}$, where $x \in G_{2N + 2}$. – Jesse Peterson Nov 9 '11 at 14:27

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