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Let $X$ be a compact (metric) space and $T:X\rightarrow X$ be a continuous map. Let $U_T:C(X)\rightarrow C(X)$ be the linear operator $U_T(f) = f\circ T$.

Then Wikipedia (see (vaguely) says that $T$ is weak topological mixing if, whenever $U_t(f)=\lambda f$ for some $\lambda\in\mathbb C$ and $f\in C(X)$, then $f$ is a constant function.

But Terry Tao (see Definition 3 of ) says that $T$ is topologically weakly mixing if $T\times T$ is topologically transitive, that is, if $U,V\subseteq X\times X$ are open then there is $n\in\mathbb Z$ with $(T\times T)^n(U)\cap V\not=\emptyset$. I guess this is equivalent to saying that for $A,B,C,D\subseteq X$ open we can find one $n\in\mathbb Z$ with both $T^n(A)\cap B$ and $T^n(C)\cap D$ non-empty.

Unfortunately, if I look in e.g. Brin and Stuck's book, then topologically transitive is defined to mean that for some $(x,y)\in X\times X$, the forward orbit $\{ (T^n(x),T^n(y)) : n\geq 1 \}$ is dense in $X\times X$.

(Edit: Thinking about this, B&S, Prop 2.2.1 shows that Tao's definition implies the B&S definition; and, at least if $T$ is a homeomorphism, the converse holds. But it doesn't seem to if $T$ is not surjective).

Is the definition of mixing involving $U_T$ equivalent to the usual one? If so, can anyone supply a reference or a sketch proof? I am beginning to think that Wikipedia has confused topological and measure-theoretic mixing.

If they are equivalent, do I really need $T$ to be a homeomorphism? (One could either replace $\mathbb Z$ by $\mathbb N$ in Tao's definition, or let $T^{-1}$ be the inverse image, and iterate). Do I need $X$ to be metric?

Edit: The reference Willie found gives the following: Suppose $T$ is a homeomorphism and $T\times T$ is minimal in the B&S sense. Then if $f\in C(X)$ with $f\circ T = \lambda f$, necessarily $\lambda\in\mathbb T$ (as $T$ is a homeomorphism). Consider $g(s,t) = f(s)\overline{f(t)}$, which defines a continuous function on $X\times X$. Then $g(T^ns,T^nt) = f(T^ns)\overline{f(T^nt)} = \lambda^n f(s) \overline{\lambda}^n \overline{f(t)} = f(s)\overline{f(t)} = g(s,t)$. There is $(x,y)$ such that $\{ (T^nx,T^ny):n\geq 1\}$ is dense in $X\times X$, from which it follows that $g$ must be constant (as $g$ is continuous). So in particular $f(x)\overline{f(y)} = |f(x)|^2$ for all $x,y$, so $f$ is constant.

So that's one implication...

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Haven't thought too much yet, but doesn't Exercise 10 in the lecture notes (stated right between Definitions 2 and 3) you linked to say precisely that the two definition of the topological transitivity agree? – Willie Wong Nov 11 '11 at 13:21
@Willie-- right you are-- but Tao's defines $T$ to be a homeomorphism, and I'd quite like to not assume that. – Matthew Daws Nov 11 '11 at 13:26
For the second part, see Theorem 2.2 and Theorem 2.3 of… which seems to say that topologically weakly mixing systems in Terry's sense (which also is the sense defined by the paper) implies weakly topologically mixing in Wikipedia's sense. – Willie Wong Nov 11 '11 at 13:35

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