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edited Nov 8 2011 at 22:04

Ian Morris
1,951●8●17

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 $1$, 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 $1$. Does When does there necessarily 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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edited Nov 8 2011 at 15:25

Ian Morris
1,951●8●17

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 $1$, 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?

Let $L$ be a bounded linear operator acting on an infinite-dimensional Banach space $B$, with the spectrum of $L$ being $\{1\}$. \{1\}$ and the norm of $L$ being $1$. Does there necessarily 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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edited Nov 8 2011 at 14:46

Ian Morris
1,951●8●17

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 $1$, 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?

Let $L$ be a bounded linear operator acting on a an infinite-dimensional Banach space $B$, with the spectrum of $L$ being $\{1\}$. Does there necessarily 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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asked Nov 8 2011 at 14:36

Ian Morris
1,951●8●17

Topological weak mixing and $\omega$-linearly-independent sequences generated by composition operators

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 $1$, 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?

Let $L$ be a bounded linear operator acting on a Banach space $B$, with the spectrum of $L$ being $\{1\}$. Does there necessarily 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!