User ian morris - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:09:19Z http://mathoverflow.net/feeds/user/1840 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131018/sequences-equidistributed-modulo-1/131036#131036 Answer by Ian Morris for Sequences equidistributed modulo 1 Ian Morris 2013-05-18T11:38:21Z 2013-05-18T12:18:16Z <p>J. F. Koksma <a href="http://www.numdam.org/item?id=CM_1935__2__250_0" rel="nofollow">proved in 1935</a> that for Lebesgue-almost-every $\beta>1$ the sequence $\beta^n$ is equidistributed modulo 1. I am not sure whether or not you would consider this to be nontrivial. Explicit examples of $\beta$ such that this property holds are not known, but it is widely thought that $\beta:=\frac{3}{2}$ has this property, as has been occasionally alluded to in answers to some <a href="http://mathoverflow.net/questions/75698/examples-of-seemingly-elementary-problems-that-are-hard-to-solve/75727#75727" rel="nofollow">other</a> <a href="http://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/102048#102048" rel="nofollow">questions</a>.</p> <p>Koksma's article also contains a range of other results of this type: for example, Satz 5 states that if $a \colon \mathbb{N} \to \mathbb{Z}$ is an arbitrary injection then $\theta \cdot a(n)$ is equidistributed modulo 1 for Lebesgue-almost-every $\theta \in (0,1)$. It is easy to then construct an example of a repetition-free sequence of integers $s_n$ such that the sequence of measures $\frac{1}{n}\sum_{i=1}^n \delta_{s_is_n^{-1} \mod 1}$ does not converge to Lebesgue measure on the interval and deduce that an appropriate constant $a>0$ exists for this sequence.</p> http://mathoverflow.net/questions/130845/blue-and-red-balls-puzzle/130940#130940 Answer by Ian Morris for Blue and red balls puzzle Ian Morris 2013-05-17T10:21:29Z 2013-05-17T13:11:38Z <p>This appears to be a description of the "OK Corral process" as an urn problem. This stochastic process was apparently introduced by David Williams and Paul McIlroy in the 1998 article <em>The OK Corral and the power of the law</em> and was subsequently investigated by J.F.C. Kingman and S. E. Volkov. In the 1999 article <em>Martingales in the OK Corral</em> Kingman proved that the expected number of balls at the end of the process is asymptotic to $n^{\frac{3}{4}}$ times $$2 \cdot 3^{-\frac{1}{4}}\pi^{-\frac{1}{2}} \Gamma\left(\frac{3}{4}\right) \simeq 1.0506511521875180068945465...$$ in the limit as $n \to \infty$. The numerical value of this constant appears to be in good agreement with Aaron's answer.</p> <p>A subsequent article by Kingman and Volkov, <a href="http://www.maths.bris.ac.uk/research/stats/reports/2001/0124.pdf" rel="nofollow"><em>Solution to the OK Corral model via decoupling of Friedman's urn</em></a>, investigates the asymptotic distribution of the probability that a particular number of balls is left at the end of the process. The authors note in particular that this process is similar to running an urn model studied by B. Friedman in 1949 in reverse time. I might add that of the three articles it is this one in which the identity between the OK Corral model and the urn model is most immediately apparent.</p> http://mathoverflow.net/questions/127799/recurrence-and-transience-of-cocycle-over-a-dynamical-system/127817#127817 Answer by Ian Morris for Recurrence and transience of cocycle over a dynamical system Ian Morris 2013-04-17T09:27:38Z 2013-04-17T09:32:50Z <p>I am not quite sure what the question is, but the following result might be a helpful starting point:</p> <p>Theorem (Giles Atkinson, 1976): Let $T$ be an ergodic invertible measure-preserving transformation of a probability space $(X,\mathcal{F},\mu)$ and let $\phi \colon X \to \mathbb{R}$ be integrable. Then the following statements are equivalent:</p> <ol> <li><p><code>$\int \phi\,d\mu=0$</code></p></li> <li><p>For all $A \in \mathcal{F}$ such that $\mu(A)>0$, and all $\varepsilon>0$, there exists a nonzero integer $n$ such that <code>$\mu\left(A \cap T^{-n}A \cap \{x \colon |\phi_n(x)|&lt;\varepsilon\}\right)&gt;0$</code>.</p></li> </ol> <p>Here we define $\phi_n(x)=0$ if $n=0$ and $\phi_n(x)=\phi_{-n}(T^nx)$ if $n$ is negative. Note that by taking $A=X$ and considering a sequence of values of $\varepsilon$ tending to zero we obtain $\liminf_{|n|\to \infty} \phi_n(x)=0$ a.e. </p> <p>This falls slightly short of proving the statement (2) in the original post since <em>a priori</em> the sequence of integers $n$ along which convergenence to zero takes place might not take infinitely many positive values. I am fairly sure that the sequence can be taken so as to tend to $+\infty$ (indeed it is not unusual to see Atkinson's theorem cited in that form) but Atkinson's original paper unfortunately does not include that statement. I seem to remember that Aaronson's book on infinite ergodic theory includes a proof and so might shed light on this question, but I appear to have mislaid my copy.</p> <p>(That a.e. $x \in X$ is recurrent in the topological sense (when $X$ is a compact metric space) when $\mu$ is an invariant Borel probability measure is a standard result: see for example Proposition 4.1.18 in Katok and Hasselblatt.)</p> http://mathoverflow.net/questions/125049/open-problems-in-pdes-dynamical-systems-mathematical-physics/125064#125064 Answer by Ian Morris for Open problems in PDEs, dynamical systems, mathematical physics Ian Morris 2013-03-20T12:49:42Z 2013-03-20T13:08:01Z <p>You could get a good overview of current research in dynamical systems via the book series <em>Handbook of Dynamical Systems</em>, which is a collection of surveys of the various areas of contemporary research in that field. Each survey is of the order of 40 pages long. This book series might also help to illustrate the sheer vastness of dynamical systems as a research topic: its four volumes total 4071 pages.</p> http://mathoverflow.net/questions/124903/liveranis-clt-a-question/124949#124949 Answer by Ian Morris for Liverani's CLT (a question) Ian Morris 2013-03-19T12:02:24Z 2013-03-19T12:19:34Z <p>Which instance of $\mathbb{E}(Y_1|\mathcal{F}_1)-0$ (or $E_1D_1=0$ in your notation) is problematic? I can find one instance shortly after (1.6) but I think that this is a typo, and the text there should read</p> <p>$\ldots$since $\mathbb{E}(\hat{T}f - \hat{T}g(\lambda) + \lambda^{-1}g(\lambda)|\mathcal{F}_1) = \mathbb{E}(Y_1(\lambda)|\mathcal{F}_1)=0$.''</p> <p>This statement can be justified by some straightforward manipulations since the sums involved now converge in $L^1$. Conversely, at this stage in the argument $g$, and hence $Y_1$, is only measurable, and I can see no reason for $\mathbb{E}(Y_1|\mathcal{F}_1)$ to even be a well-defined expression.</p> <p>Once one has established $\mathbb{E}(Y_1^2)&lt;\infty$, it is then implicitly necessary to go back and show that indeed $\mathbb{E}(Y_1|\mathcal{F}_1)=0$. Is this the problematic step? (I am certainly not quite sure at this stage how to do it.)</p> http://mathoverflow.net/questions/124615/equivalence-of-two-definitions-of-lyapunov-exponents/124713#124713 Answer by Ian Morris for Equivalence of two definitions of Lyapunov exponents Ian Morris 2013-03-16T18:17:49Z 2013-03-16T18:17:49Z <p>To expand on Anthony's comment somewhat, the two definitions differ in that the second limit can fail to exist even when the first limit does exist. For example, consider a case in which the sequence of Jacobian matrices alternates between the following two matrices: $$A_1:=\left(\begin{array}{cc}0&amp;2\\2&amp;0\end{array}\right),\qquad A_2:=\left(\begin{array}{cc}0&amp;1\\4&amp;0\end{array}\right).$$ (This could arise if, for example, $f$ is a transformation of the disjoint union of two 2-tori which acts by first applying to each torus the toral automorphism defined by one of the above matrices, and then subsequently interchanging the two tori.) Now, products of the form $(A_2A_1)^n$ or $(A_1A_2)^n$ have $\lambda_1=8^n$ and $\lambda_2=2^n$, but products of the form $A_1(A_2A_1)^n$ or $A_2(A_1A_2)^n$ have $\lambda_1=\lambda_2=2^{2n+1}$. The second limit therefore fails to exist. On the other hand, RW's condition (*) is clearly satisfied, and it is not difficult to show that the first and second singular values which appear in the first definition are respectively $8^n$ and $2^n$ for all of the above products and so the first limit exists.</p> http://mathoverflow.net/questions/124066/non-existence-of-ergodic-measures/124077#124077 Answer by Ian Morris for Non-existence of ergodic measures Ian Morris 2013-03-09T13:26:19Z 2013-03-09T17:08:30Z <p>The existence of such an example is prevented by the ergodic decomposition theorem, which asserts that every $T$-invariant measure on a standard probability space $(X,\mathcal{B},m)$ can be expressed as a (possibly uncountably infinite) convex combination of ergodic $T$-invariant measures by means of an integral over the set of ergodic $T$-invariant measures on $(X,\mathcal{B},m)$. In particular, since the integral must have a nonzero outcome the set of such measures is nonempty.</p> <p>This theorem is relatively technical to prove and seems to be left unproved or even unstated in standard textbooks on ergodic theory. For example, Walters (p.34) describes the theorem without stating it formally, referring instead to the original work of V. I. Rokhlin (<em>Selected topics in the metric theory of dynamical systems</em>, Uspekhi Mat. Nauk. <strong>4</strong> (1949) 58--127). Petersen (p.81) mentions the result in passing but does not provide a reference. Aaronson's book on infinite ergodic theory gives a proof of the ergodic decomposition theorem for probability spaces in the case where $T$ is invertible (p. 62--64) and sets the case of general $T$ as an exercise; Aaronson attributes the result to von Neumann (who discovered it independently of Rokhlin) and cites <em>Zur Operatorenmethode in der Klassischen Mechanik</em>, Ann. Math. <strong>33</strong> (1932) 587--642.</p> <p>Terence Tao's weblog has a nice discussion of the result <a href="http://terrytao.wordpress.com/2008/02/04/254a-lecture-9-ergodicity/#more-252" rel="nofollow">here</a>.</p> http://mathoverflow.net/questions/122764/characterising-ergodicity-of-continuous-maps/122769#122769 Answer by Ian Morris for Characterising ergodicity of continuous maps Ian Morris 2013-02-24T01:35:38Z 2013-02-24T01:35:38Z <p>Let $T \colon X \to X$ be a minimal transformation of a compact metric space which is not uniquely ergodic, let $\mu$ be a non-ergodic $T$-invariant measure on $X$, and let $A$ be a set with nonempty interior such that $\mu(A \triangle T^{-1}A)=0$. I claim that necessarily $\mu(A)=1$, contradicting the above conjecture. (Some constructions of transformations with the above combination of properties may be found for example in the textbook <em>Ergodic Theory on Compact Spaces</em> by Denker, Grillenberger and Sigmund, or in John Oxtoby's classic 1952 article <em>Ergodic sets</em>.)</p> <p>Let $U \subseteq A$ be open and nonempty. Since $T$ is minimal we have $\bigcup_{n=0}^\infty T^{-n}U=X$, and indeed even $\bigcup_{n=0}^NT^{-n}U=X$ for some integer $N$ since $X$ is compact. In particular $\bigcup_{n=0}^N T^{-n}A=X$. Let us write $$\bigcup_{n=0}^N T^{-n}A = A \cup \bigcup_{n=1}^N \left(\left( T^{-n}A\right)\setminus \bigcup_{k=0}^{n-1} T^{-k}A\right)=A \cup \bigcup_{n=1}^N B_n,$$ say, which is a disjoint union. We would like to show that this union has measure identical to that of $A$. For each $n$ we have $$\mu(B_n)=\mu\left(T^{-n}A\setminus \bigcup_{k=0}^{n-1} T^{-k}A\right)\leq \mu\left(T^{-n}A \setminus T^{-(n-1)}A\right)=\mu\left(T^{-1}A \setminus A\right)=0$$ by invariance and the hypothesis $\mu(A \triangle T^{-1}A)=0$. It follows that $$\mu(A)=\mu\left(\bigcup_{n=0}^N T^{-n}A \right)=\mu(X)=1$$ so the desired situation can not occur.</p> http://mathoverflow.net/questions/117415/old-books-still-used/117502#117502 Answer by Ian Morris for Old books still used Ian Morris 2012-12-29T10:54:26Z 2012-12-29T11:00:31Z <p>My own field, ergodic theory, is relatively young in that some concepts now regarded as fundamental -- Kolmogorov-Sinai entropy, for example -- were not fully formulated until around 1960. Nonetheless there are a couple of old books still in use and receiving citations:</p> <p>E. Hopf, <em>Ergodentheorie</em>, 1937;</p> <p>R. Halmos, <em>Ergodic theory</em>, 1957.</p> <p>If the 1960s are sufficiently long ago to constitute "old" then there are many old references in probability which remain in heavy use, for example:</p> <p>P. Billingsley, <em>Convergence of probability measures</em>, 1968;</p> <p>L. Breiman, <em>Probability</em>, 1968;</p> <p>and one of the classics of the field:</p> <p>W. Feller, <em>Introduction to probability theory and its applications</em>, 1950.</p> <p>Outside my own field, a much-cited number theory text which no-one has yet mentioned:</p> <p>A. Khinchin, <em>Continued fractions</em>, 1936.</p> http://mathoverflow.net/questions/117203/estimate-entropy-of-a-binary-process-in-terms-of-decay-of-correlations/117209#117209 Answer by Ian Morris for Estimate entropy of a binary process in terms of decay of correlations Ian Morris 2012-12-25T20:14:28Z 2012-12-25T20:14:28Z <p>I would not be surprised if small $\sigma^2$ does indeed imply small d-bar distance to Bernoulli, but I think that the converse is false. Rather, processes with arbitrarily small d-bar distance to Bernoulli can have $\sigma^2=+\infty$.</p> <p>To see this let $(x_n)$ be a sequence which generates the $(\frac{1}{2},\frac{1}{2})$-Bernoulli process, and let $N>0$. Define a new sequence $(z_n)$ by taking $z_n:=x_n$ when $n$ is not divisible by $2^N$. Otherwise, take $z_n:=1$ if $n=k2^N$ for even $k$, and $z_n:=0$ if $n=k2^N$ for odd $k$.</p> <p>Now consider the (ergodic) process generated by $(z_n)$. It is not difficult to see that $0$ and $1$ have equal probability, and when $n$ is divisible by $2^{N+1}$ we have $$\mathbb{P}(X_1=X_n=1)=\left(1-\frac{1}{2^N}\right)\cdot \left(\frac{1}{2}\right) + \left(\frac{1}{2^N}\right)\cdot 1 = \frac{1}{2}+\frac{1}{2^{N+1}}$$ so that $$\sup_{a,b}\left|\mathbb{P}(X_1=a,X_n=b)-\mathbb{P}(X_1=a)\mathbb{P}(X_n=b)\right|\geq \frac{1}{2^{N+1}}.$$ In particular $\sigma^2=+\infty$ as claimed.</p> http://mathoverflow.net/questions/111948/modern-version-of-an-inequality-of-r-m-gabriel-for-contour-integrals Modern version of an inequality of R. M. Gabriel for contour integrals Ian Morris 2012-11-09T22:54:20Z 2012-11-30T19:49:27Z <p>I am currently reading the 1998 article <em>Dynamics of the Binary Euclidean Algorithm: Functional Analysis and Operators</em> by Brigitte Vallée, which cites a 1928 article by R. M. Gabriel for the following inequality: if $f$ is a holomorphic function defined in an open ball $U$, $\Gamma$ is a circular contour contained in $U$, $\gamma$ is a convex contour enclosed by $\Gamma$, and $p>0$, then $$\int_\gamma |f(z)|^p|dz|\leq 2\int_\Gamma |f(z)|^p|dz|.$$ The text suggests that Vallée borrowed this reference from the 1969 PhD thesis of Howard J. Schwartz - which deals with composition operators on Hardy spaces - although I haven't seen this thesis myself. I found Gabriel's original paper somewhat dated in its terminology and presentation, and I am wondering whether a more up-to-date reference exists for this result, or whether it has been subsumed into a more general result which is now relatively well-known. Is anyone able to point me in the direction of a modern version of this inequality, or a textbook which contains this inequality? In Vallée's application and in the one which I am considering, it is sufficient to consider the case in which $\gamma$ is also circular.</p> <p>Since the result holds for all $p>0$, I wonder whether the key property being used is subharmonicity rather than holomorphicity. In any case, I haven't been able to find this result either in books on Hardy spaces (the context in which it is applied by Vallée, and presumably also Schwartz) or in books on convex analysis; or if I have found it, its modern form is so far removed from Gabriel's statement that I was unable to recognise it. Anyway, I would be very grateful if someone could direct me to a more modern reference for this result.</p> http://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/102048#102048 Answer by Ian Morris for Not especially famous, long-open problems which anyone can understand Ian Morris 2012-07-12T16:03:29Z 2012-07-25T11:11:34Z <p>Is the sequence $(3/2)^n \mod 1$ dense in the unit interval? It is known that $\beta^n$ is uniformly distributed modulo one for almost all $\beta>1$, but explicit examples of $\beta$ for which density holds are not known. This question seems to originate in work of Weyl and Koksma on uniform distribution.</p> <p>Update: Since posting this answer I've attempted to find some references with which to flesh it out, with only modest success. The earlier paper I have identified which deals with this question directly is T. Vijayaraghavan's 1940 article <em>On the fractional parts of the powers of a number</em>, in which it is shown that the sequence $(3/2)^n \mod 1$ has infinitely many limit points. Jeffrey Lagarias' 1985 survey on the Collatz problem, <em>The 3x + 1 Problem and Its Generalizations</em>, includes a one-page overview of the literature on the distribution of this sequence. Flatto, Lagarias and Pollington subsequently proved that the diameter of the set of accumulation points is at least 1/3.</p> http://mathoverflow.net/questions/101859/a-fourier-analytic-inequality-used-by-jean-bourgain A Fourier-analytic inequality used by Jean Bourgain Ian Morris 2012-07-10T15:09:40Z 2012-07-10T20:48:36Z <p>I am currently reading Jean Bourgain's 1986 paper <i>A Szemerédi type theorem for sets of positive density in</i> $R^k$ and would appreciate some help in understanding a Fourier-analytic estimate used in that article. I suspect that my question is relatively elementary, but my knowledge of Fourier analysis is not very strong and there are no experts in the topic at my current place of work, so I would very much appreciate a pointer.</p> <p>In Bourgain's argument, $f \colon \mathbb{R}^d \to [0,1]$ is a nonzero measurable function supported in a fixed bounded measurable set $A$, and the $L^2$ norm of $f$ is fixed. For each $\lambda>0$ we define $P_\lambda \colon \mathbb{R}^d \to \mathbb{R}$ to be the function whose Fourier transform <code>$\hat{P}_\lambda(\xi):=\int_{\mathbb{R}^d} e^{-2\pi i \langle x,\xi\rangle} P_\lambda(x)dx$</code> is given by $\hat{P}_\lambda(\xi)=e^{-\lambda\|\xi\|}$ for all $\xi \in \mathbb{R}^d$. Parameters $\delta, t>0$ are introduced, and the parameter $\delta$ is subsequently fixed at some small value which depends on $\|f\|_2$ (and possibly on $A$) but not on the precise choice of $f$. It is then claimed that by taking $t$ small enough, the quantity $$\|(f * P_{\delta t}) - (f * P_{\delta^{-1}t})\|_2$$ can be made arbitarily small in a manner which is uniform with respect to $f$. It is clear to me that this quantity must converge to zero as $t \to 0$ when $\delta$ and $f$ are fixed, but it is not clear to me why a single value $t$ can be chosen which works simultaneously for all $f$ (where $\|f\|_2$ is fixed and the support of $f$ lies in $A$). Bourgain's paper seems to use a quantitative bound which I infer to resemble $$\|(f * P_{\delta t}) - (f * P_{\delta^{-1}t})\|_2 \leq C\|f\|_2\frac{\log (1/\delta)}{\log (1/t)}.$$ Certainly it is stated that in order to make the above difference small (relative to $\delta^{1/4}$ and $\|f\|_2$) it is sufficient that $\log (1/t)$ should be a large multiple of $\log (1/\delta)$. Can anyone see more precisely what estimate is being used here, or at least how the above quantity can be bounded uniformly with respect to $f$?</p> <p>Thanks!</p> http://mathoverflow.net/questions/85984/density-of-strictly-ergodic-measures-in-the-d-bar-topology Density of strictly ergodic measures in the d-bar topology Ian Morris 2012-01-18T12:19:51Z 2012-01-18T12:19:51Z <p>I am currently studying a problem which deals with cocycles of highly noncompact operators on Hilbert space, with the base transformation being the full shift on two symbols. In my particular situation it turns out that the top Lyapunov exponent of a fixed cocycle, considered as a function on the space of invariant measures, is discontinuous with respect to the weak-* topology but Lipschitz continuous with respect to Ornstein's $\overline{d}$-metric. However I do not have very much intuition for the topology on the invariant measures induced by $\overline{d}$, and the resources on this seem relatively limited (the best I have found so far is Glasner's book <em>Ergodic theory via joinings</em>). I am currently trying to understand the generic features of the set of ergodic measures with respect to the $\overline{d}$-metric.</p> <p>With regard to the space of shift-invariant measures under the weak-* topology, the following facts have been well-known for many years, mostly dating back to Parthasarathy's 1961 paper <em>On the category of ergodic measures</em>:</p> <ul> <li>The space of measures under the weak-* topology is a compact metrisable space.</li> <li>Measures supported on a single periodic orbit are dense. In particular, ergodic measures are dense, and strictly ergodic measures (i.e. those whose support is uniquely ergodic) are dense.</li> <li>Weak-mixing measures are a dense residual set.</li> <li>Strong-mixing measures are a dense meagre set.</li> <li>Fully-supported measures are a dense residual set.</li> <li>Zero-entropy measures are a dense residual set.</li> </ul> <p>Most of these statements have known analogues in the $\overline{d}$-metric, namely:</p> <ul> <li>The space of measures under the $\overline{d}$-metric is complete but not separable.</li> <li>Measures supported on periodic orbits are not dense.</li> <li>The space of ergodic measures is closed, as are the spaces of strong-mixing and Bernoulli measures.</li> <li>Fully-supported measures are a dense residual subset of the ergodic measures.</li> <li>Entropy is continuous.</li> </ul> <p>However, it is not clear to me whether or not strictly ergodic measures are dense in the ergodic measures in the $\overline{d}$-metric. Does anyone know whether or not this is the case?</p> http://mathoverflow.net/questions/71683/alternative-proofs-of-the-krylov-bogolioubov-theorem Alternative proofs of the Krylov-Bogolioubov theorem Ian Morris 2011-07-30T20:29:11Z 2012-01-09T19:00:59Z <p>The Krylov-Bogolioubov theorem is a fundamental result in the ergodic theory of dynamical systems which is typically stated as follows: if $T$ is a continuous transformation of a nonempty compact metric space $X$, then there exists a Borel probability measure $\mu$ on $X$ which is invariant under $T$ in the sense that $\mu(A)=\mu(T^{-1}A)$ for all Borel sets $A \subseteq X$, or equivalently $\int f d\mu = \int (f \circ T)d\mu$ for every continuous function $f \colon X \to \mathbb{R}$. This question is concerned with proofs of that theorem.</p> <p>The most popular proof of the Krylov-Bogolioubov theorem operates as follows. Let $\mathcal{M}$ denote the set of all Borel probability measures on $X$, and equip $\mathcal{M}$ with the coarsest topology such that for every continuous $f \colon X \to \mathbb{R}$, the function from $\mathcal{M}$ to $\mathbb{R}$ defined by $\mu \mapsto \int f d\mu$ is continuous. In this topology $\mathcal{M}$ is compact and metrisable, and a sequence $(\mu_n)$ of elements of $\mathcal{M}$ converges to a limit $\mu$ if and only if $\int fd\mu_n \to \int fd\mu$ for every continuous $f \colon X \to \mathbb{R}$. Now let $x \in X$ be arbitrary, and define a sequence of elements of $\mathcal{M}$ by $\mu_n:=(1/n)\sum_{i=0}^{n-1}\delta_{T^ix}$, where $\delta_z$ denotes the Dirac probability measure concentrated at $z$. Using the sequential compactness of $\mathcal{M}$ we may extract an accumulation point $\mu$ which is invariant under $T$ by an easy calculation. This proof, together with minor variations thereupon, is fairly ubiquitous in ergodic theory textbooks. In the answers to <a href="http://mathoverflow.net/questions/66669/proof-of-krylov-bogoliubov-theorem" rel="nofollow">this question</a>, Vaughn Climenhaga notes the following alternative proof: the map taking the measure $\mu$ to the measure <code>$T_*\mu$</code> defined by <code>$(T_*\mu)(A):=\mu(T^{-1}A)$</code> is a continuous transformation of the compact convex set $\mathcal{M}$, and hence has a fixed point by the Schauder-Tychonoff theorem. A couple of years ago I thought of another proof, given below. The first part of this question is: has the following proof ever been published?</p> <p>This third proof is as follows. Clearly it suffices to show that there exists a <em>finite</em> Borel measure on $X$ which is invariant under $T$, since we may normalise this measure to produce a probability measure. By the Hahn decomposition theorem it follows that it suffices to find a nonzero finite <em>signed</em> measure on $X$ which is invariant under $T$. By the Riesz representation theorem for measures this is equivalent to the statement that there exists a nonzero continuous linear functional $L \colon C(X) \to \mathbb{R}$ such that $L(f \circ T)=f$ for all continuous functions $f \colon X \to \mathbb{R}$. Let $B(X)$ be the closed subspace of $C(X)$ which is equal to the closure of the set of all continuous functions which take the form $g \circ T - g$ for some continuous $g$. Clearly a continuous linear functional $L \colon C(X) \to \mathbb{R}$ satisfies $L(f \circ T)=f$ if and only if it vanishes on $B(X)$, so to construct an invariant measure it suffices to show that the dual of $C(X)/B(X)$ is nontrivial. A consequence of the Hahn-Banach theorem is that the dual of $C(X)/B(X)$ is nontrivial as long as $C(X)/B(X)$ is itself nontrivial, so to prove the theorem it is sufficient to show that the complement of $B(X)$ in $C(X)$ is nonempty. But the constant function $h(x):=1$ is not in $B(X)$ because if $|(g \circ T - g)(x) - 1|&lt;1/2$ for all $x \in X$, then $g(Tx)>g(x)+1/2$ for all $x \in X$ and hence $g(T^nx)>g(x)+n/2$ for all $n \geq 1$ and $x \in X$. This is impossible since $g$ is continuous and $X$ is compact. We conclude that $B(X)$ is a proper Banach subspace of $C(X)$ and the desired functional exists.</p> <p>The second part of the question deals with the Krylov-Bogolioubov theorem for measures invariant under amenable groups of transformations. Let $\Gamma = \{T_\gamma\}$ be a countable amenable group of continuous transformations of the compact metric space $X$. We shall say that $\mu \in \mathcal{M}$ is invariant under $\Gamma$ if $(T_\gamma)_*\mu = \mu$ for all $T_\gamma$. I believe that by using Følner sequences one may generalise the first proof of the Krylov-Bogolioubov theorem to show that every such amenable group has an invariant Borel probability measure. I seem to recall that the second proof also generalises to this scenario (in Glasner's book, perhaps?). Despite a certain amount of thought I have not been able to see how the third proof might generalise to this situation, even when $\Gamma$ is generated by just two commuting elements. So, the second part of this question is: can anyone see how the third proof generalises to the amenable case?</p> http://mathoverflow.net/questions/81541/fourier-transform-of-x2-invariant-measure/81601#81601 Answer by Ian Morris for Fourier transform of x2 invariant measure Ian Morris 2011-11-22T12:10:00Z 2011-11-22T12:42:55Z <p>As Andreas Thom correctly points out, any invariant Borel probability measure $\hat\mu$ must satisfy $\hat\mu(n)=\hat\mu(2n)$ for all $n \in \mathbb{Z}$, so if the Fourier coefficients tend to zero in the limit then all of them except $\hat\mu(0)$ must be identically zero.</p> <p>There are at least some additional constraints on the limiting behaviour of the Fourier coefficients of an invariant measure. Since Lebesgue measure is an ergodic measure for $T$, there can be no other invariant measures which are absolutely continuous with respect to Lebesgue. (This can be proved either by showing that the density must be invariant and therefore constant, or by using the Birkhoff ergodic theorem to obtain a contradiction.) While this clearly further constrains the behaviour of the Fourier coefficients, my grasp of Fourier analysis isn't strong enough for me to be able to describe the effect on the coefficients in exact terms.</p> <p>Any general statement about the behaviour of Fourier coefficients for positive-entropy invariant measures of the doubling map will be constrained by the fact that the set of all such measures is extremely large: every ergodic measure-preserving transformation of a probability space with entropy less than $\log 2$ can be realised as the doubling map equipped with some invariant measure.</p> http://mathoverflow.net/questions/80687/is-it-realistic-to-want-to-classify-minimal-sub-systems-in-small-dimension/80754#80754 Answer by Ian Morris for Is it realistic to want to classify minimal sub-systems (in small dimension) ? Ian Morris 2011-11-12T14:01:06Z 2011-11-12T14:39:48Z <p>It is certainly the case that classifying the minimal subsystems of homeomorphisms of compact 2-manifolds presents profound and fundamental difficulties. This is because some very simple transformations, such as analytic diffeomorphisms of the 2-torus, have extremely rich families of minimal sets. </p> <p>Let $T \colon X \to X$ be a linear Anosov diffeomorphism of the 2-torus. The topological entropy of $T$ is finite and positive but may be arbitrarily large. If a natural number $k$ is specified, then we may find a linear Anosov diffeomorphism $T$ of the 2-torus $X$ such that the shift transformation on $k$ symbols may be homeomorphically embedded into the dynamical system $(X,T)$ as a compact invariant subset. In particular, every minimal subsystem of the $k$-shift embeds into $(X,T)$ as a minimal subsystem. </p> <p>This is problematic because the $k$-shift has an enormous number of minimal subsystems, all of which will be inherited by the Anosov system. Indeed, the combinatorial version of the Jewett-Krieger theorem implies that every ergodic measure-preserving transformation of an abstract probability space which has entropy strictly less than $\log k$ may be embedded into the $k$-shift as a uniquely ergodic minimal subsystem. In particular, for a linear Anosov diffeomorphism of the 2-torus with topological entropy large enough, every ergodic measurable dynamical system with measure-theoretic entropy up to some threshold arises as a uniquely ergodic minimal subsystem.</p> <p>This already presents us with an enormous number of minimal subsystems, because for each $h \geq 0$ there exist uncountably many ergodic measurable dynamical systems of entropy $h$ which are not pairwise equivalent. This is then compounded by the fact that some minimal systems of $(X,T)$ will not arise from such an embedding, the fact that the embedding of the abstract ergodic system into the $k$-shift is in general not unique, the fact that the embedding of the $k$-shift into $(X,T)$ is in general not unique, and the fact that the $k$-shift itself has additional minimal subsystems. Indeed, there is a further theorem due to Denker, Grillenberger and Sigmund which implies that for any finite collection of abstract ergodic transformations all having entropy strictly below $\log k$, we can find a minimal subsystem of the $k$-shift which has an embedded copy of each of these transformations as its only ergodic measures.</p> <p>On the basis of the above considerations I think that a satisfactory classification of the minimal subsystems of homeomorphisms of the 2-torus is improbable!</p> http://mathoverflow.net/questions/80391/topological-weak-mixing-and-omega-linearly-independent-sequences-generated-by Topological weak mixing and $\omega$-linearly-independent sequences generated by composition operators Ian Morris 2011-11-08T14:36:10Z 2011-11-09T07:11:30Z <p>A research problem on which I am currently working requires a construction in topological dynamics of the following type:</p> <blockquote> <p>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?</p> </blockquote> <p>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 <em>topological weak mixing</em>. Specifically, I ask:</p> <blockquote> <p>Is topological weak mixing of $T$ sufficient to guarantee the existence of $f$ in the first question?</p> </blockquote> <p>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:</p> <blockquote> <p>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?</p> </blockquote> <p>Thanks in advance!</p> http://mathoverflow.net/questions/75131/how-to-detect-frequency/75139#75139 Answer by Ian Morris for How to detect frequency? Ian Morris 2011-09-11T11:41:07Z 2011-09-11T11:41:07Z <p>It seems likely to me that $\alpha$ can be computed by calculating the frequencies of subwords of the coding sequence, but in a manner which depends on certain parameters. For example, if $\alpha&lt;\min\{|J|,2\pi-|J|\}$ then the interval $J \setminus J +\alpha$ has length precisely $\alpha$, and it follows easily that $\alpha$ equals the frequency of the subword 01. On the other hand if $|J|$ is very small and $\alpha, 2\pi-\alpha$ are both larger than $|J|$, then the frequency of the subwords 01 and 10 are both $|J|$, while the subword 00 has frequency $1-2|J|$, and we cannot gain anything by considering words of length 1 or 2. So the frequencies of words of arbitrary length probably need to be considered.</p> <p>The articles "Coding rotations on intervals" by Berstel and Vuillon, and "Three-distance theorems and combinatorics on words" by Alessandri and Berthé appear to be relevant (especially Lemma 1 in the latter) but do not seem to yield a complete answer.</p> http://mathoverflow.net/questions/73248/applications-of-and-motivation-for-von-neumanns-mean-ergodic-theorem/73276#73276 Answer by Ian Morris for Applications of and motivation for von Neumann's mean ergodic theorem Ian Morris 2011-08-20T13:10:16Z 2011-08-20T13:22:20Z <p>I believe that Hillel Furstenberg uses the von Neumann ergodic theorem quite frequently in his work on recurrence, which has applications to number theory. For example, in section 3 of the article <em>Poincaré recurrence and number theory</em> he uses Weyl's criterion and the von Neumann ergodic theorem to prove the following result: if $T$ is a measure-preserving transformation of a probability space, $p$ is a polynomial with integer coefficients and no constant term, and $A$ is a positive-measure subset of the probability space in question, then there are infinitely many natural numbers $t$ such that $T^{-p(t)}A \cap A$ has postive measure. A corollary of this result is that if $X$ is a subset of the integers with positive density and $p$ is an integer polynomial with no constant term, then the equation $x-y=p(t)$ can be solved for $x,y \in X$ and $t$ a positive integer. The von Neumann ergodic theorem is also used in ergodic proofs of Roth's theorem (see for example the exposition by Á. Magyar). There are probably more examples in the book <em>Recurrence in Ergodic Theory and Combinatorial Number Theory</em>.</p> http://mathoverflow.net/questions/71636/a-follow-up-question-related-to-entropy/71669#71669 Answer by Ian Morris for A follow up question related to entropy Ian Morris 2011-07-30T17:52:55Z 2011-07-30T18:05:27Z <p>For the topological entropy of a subshift on finitely many symbols, I think that this limit will typically be infinite. Here is an example where this is the case.</p> <p>Let $\Sigma_2= \{0,1\}^{\mathbb{N}}$ with the infinite product topology and let $T \colon \Sigma_2 \to \Sigma_2$ denote the shift transformation given by <code>$T[(x_i)_{i=1}^\infty]:=(x_{i+1})_{i=1}^\infty$</code>. The map $T$ is a continuous surjective transformation of the compact metrisable space $\Sigma_2$. Let us define a <em>cylinder set</em> of depth $n$ to be a set $Z \subseteq \Sigma_2$ having the form $$Z=\{(x_i) \in \Sigma_2 \colon x_j=z_j \text{ for all }j\text{ such that }1 \leq j \leq n\}$$ for some finite sequence of symbols $z_1,\ldots,z_n \in \{0,1\}$. If $K$ is a nonempty compact subset of $\Sigma_2$ such that $TK \subseteq K$, then the topological entropy of $T$ restricted to $K$ admits the following description: if for each $n \geq 1$ we let $a_n$ be the number of distinct cylinder sets of depth $n$ which have nonempty intersection with $K$, then <code>$h_{top}(K) = \lim_{n \to \infty} \frac{1}{n} \log a_n$</code>. This holds because the cylinder sets form the smallest-growing family of open covers in the sense of the usual definition of topological entropy on a compact space.</p> <p>Now, let $K \subset \Sigma_2$ be a compact $T$-invariant set of Sturmian words with some specified irrational slope (for the definition and fundamental properties of Sturmian words see e.g. Wikipedia). Such sets exist and satisfy $a_n=n+1$ for all $n \geq 1$. In particular the restriction of the shift map $T$ to $K$ has topological entropy zero and $a_ne^{-nh}=n+1 \to \infty$.</p> <p>More generally, a little further thought shows that $a_ne^{-nh} \to \infty$ for every nonempty compact minimal invariant subset of $\Sigma_2$ which has zero topological entropy and is not equal to a periodic orbit.</p> http://mathoverflow.net/questions/70676/non-oscillatory-behaviour-in-the-subadditive-ergodic-theorem Non-oscillatory behaviour in the subadditive ergodic theorem Ian Morris 2011-07-18T21:27:14Z 2011-07-19T05:05:37Z <p>I am currently reading an article in which the author goes to certain lengths which could be avoided if the following result were true:</p> <blockquote> <blockquote> <p><em>Lemma (proposed)</em>: Let $T$ be an ergodic measure-preserving transformation of a probability space $(X,\mathcal{F},\mu)$, and let $(f_n)$ be a sequence of integrable functions from $X$ to $\mathbb{R}$ which satisfy the subadditivity relation $f_{n+m} \leq f_n \circ T^m + f_m$ a.e. for all integers $n,m \geq 1$. Suppose that $f_n(x) \to -\infty$ in the limit as $n \to \infty$ for $\mu$-a.e. $x \in X$. Then $\lim_{n \to \infty} \frac{1}{n}\int f_n d\mu &lt;0$.</p> </blockquote> </blockquote> <p>Via the subadditive ergodic theorem, this effectively states that if $f_n(x) \to -\infty$ almost everywhere then it must do so at an asymptotically linear rate. The supposed lemma would also be equivalent to the statement that if $\frac{1}{n} f_n(x) \to 0$ almost everywhere, then for almost every $x$ the sequence $(f_n(x))$ must return infinitely often to some neighbourhood of $0$ which is not a neighbourhood of $-\infty$. If the sequence $(f_n)$ is additive rather than just subadditive then this last formulation of the result follows from a well-known theorem of G. Atkinson, but the more general subadditive case is less clear.</p> <p>If the lemma were true then several parts of the paper I am reading would be redundant, which makes me wonder whether it is in fact false. Yet it seems rather plausible. Does anyone know whether this result is true or not?</p> http://mathoverflow.net/questions/59222/importance-of-poincare-recurrence-theorem-any-example/59241#59241 Answer by Ian Morris for Importance of Poincaré recurrence theorem? Any example? Ian Morris 2011-03-22T22:16:43Z 2011-03-23T10:55:30Z <p>The Poincaré recurrence theorem is sometimes useful because of the way it translates into recurrence in metric spaces. For example, a corollary of the Poincaré theorem is that for a measure-preserving transformation of a separable metric space - which need not be continuous - almost every point is recurrent in the topological sense. To see this, choose a sequence which is dense in the metric space, and consider the cover of the metric space by balls of radius $1/n$ around points in this sequence. By the Poincaré theorem almost every point belonging to one of these balls returns to that ball infinitely often, and hence returns to within distance $2/n$ of itself. Now take the intersection over $n$ to see that almost every point is recurrent.</p> <p>A more general corollary which is also sometimes useful is the following: if $T \colon X \to X$ is measure-preserving, and $f$ is a measurable function from $X$ to a separable metric space, then $\liminf_{n \to \infty} d(f(T^nx),f(x))=0$ almost everywhere. This result naturally can be very useful in circumstances where one knows that a function is measurable, but no more. An example which springs to mind is the proof of Theorem 15 in "A formula with some applications to the theory of Lyapunov exponents" by Avila and Bochi, where the Poincaré theorem is applied to prove the recurrence of the measurable splittings in the multiplicative ergodic theorem. </p> http://mathoverflow.net/questions/56215/generic-points-and-local-entropies/56274#56274 Answer by Ian Morris for Generic points and local entropies Ian Morris 2011-02-22T12:32:13Z 2011-02-22T12:32:13Z <p>My feeling is that there exists an ergodic measure $\mu$ for which $G_\mu \setminus Z_\mu$ is nonempty. It is sufficient to find a uniquely ergodic subsystem which admits exceptional points for the Shannon-McMillan-Breiman theorem. I think that one can be constructed symbolically without too much difficulty by the following method.</p> <p>Pick a real number $h$ lying strictly between 0 and $\log 2$, and consider a sequence $x$ in the 2-shift with the following properties:</p> <p>1) For every $n \geq 1$, the sequence contains precisely $e^{nh + o(n)}$ distinct words of length $n$. (For reasons of subadditivity the $o(n)$ term is necessarily positive).</p> <p>2) Every word which occurs in $x$ occurs with a well-defined frequency which is not equal to 0 or 1.</p> <p>The orbit closure $X$ of such a sequence is then a uniquely ergodic subsystem of the shift with topological entropy equal to $h$. An explicit procedure for constructing such a sequence was given by Grillenberger in the 1970s (in my opinion it's not particularly hard). In particular, $X$ supports a unique invariant measure $\mu$ and $G_\mu$ includes the whole of $X$. Now, suppose that the word $x$ also satisfies the property:</p> <p>3) There exists a nested sequence of subwords of $x$ such that the frequency of each of these words is less than $e^{-n(h+\varepsilon)}$ for some $\varepsilon>0$.</p> <p>This implies that there is a nested sequence of cylinder sets in $X$, containing some point, such that the measures of these cylinder sets decrease at a rate faster than the "standard" local entropy $h$, and hence the point in the intersection of the cylinders belongs to $G_\mu$ but not to $Z_\mu$.</p> <p>I think that there shouldn't be any problem in reconciling all three of these criteria with one another, but I will admit that I haven't attempted to write a proof of that. I think it sounds reasonable that for a larger class of measures than Gibbs measures we should have $G_\mu \subseteq Z_\mu$, but I don't have much to contribute to that end of the question...</p> http://mathoverflow.net/questions/8056/what-are-good-non-english-languages-for-mathematicians-to-know/8114#8114 Answer by Ian Morris for What are good non-English languages for mathematicians to know? Ian Morris 2009-12-07T17:13:28Z 2011-02-01T10:16:47Z <p>A quick examination suggests that articles in French continue to appear in <em>Annals</em> and <em>Inventiones</em> with some regularity, though much more frequently in the European <em>Inventiones</em> than the US-based <em>Annals</em>. I think this quite strongly supports the position that French is a useful language to be able to read. On the other hand, neither of these two journals seems to have published anything in German, or indeed any language other than French or English, for quite some time. Germany's other major journals, <em>Mathematische Annalen</em> and <em>Journal für die reine und angewandte Mathematik</em>, also do not these days seem to publish articles in German in practice. Russian and Chinese are certainly very active mathematical languages in their countries of origin, but unlike French, major works in these languages seem typically much more subject to translation.</p> http://mathoverflow.net/questions/51563/topology-on-the-set-of-linear-subspaces/51639#51639 Answer by Ian Morris for Topology on the set of linear subspaces Ian Morris 2011-01-10T11:04:11Z 2011-01-10T11:04:11Z <p>Some of the answers to <a href="http://mathoverflow.net/questions/48118" rel="nofollow">this question</a> might be helpful for your question also. It deals with finite-dimensional Hilbert spaces, but most of my answer to that question applies to the infinite-dimensional case too, with one or two obvious exceptions (e.g. the metric space of 1-dimensional subspaces of an infinite-dimensional Hilbert space is not compact). In particular, the book on Hilbert spaces by Akhiezer and Glazman has a short (5 pages?) section on the Grassmannian of a Hilbert space, and shows that the metric on the Grassmannian given by `aperture' is the same as the metric given by the operator difference between orthogonal projections.</p> http://mathoverflow.net/questions/48118/a-metric-for-grassmannians/48440#48440 Answer by Ian Morris for A metric for Grassmannians Ian Morris 2010-12-06T13:33:33Z 2010-12-06T19:18:51Z <p>I found it surprisingly difficult to find a reference for this when I was studying Mane's papers on multiplicative ergodic theorems. My hypothesis was that people working with the Grassmannian in other areas are happy with the fact that the Grassmannian is metrisable for abstract topological reasons, and don't actually care very much about a precise metric, but I might be wrong about this... in my answer I'm going to assume that we're considering a finite-dimensional space equipped with an inner product structure.</p> <p>If you are interested in precise metrics on the Grassmannian, the most popular definition of which I am aware is this one: <code>$$d(V,W):=\max\left\{\sup_{w \in W, \|w\|=1}\inf \{\|v-w\| \colon v \in V \},\sup_{v \in V, \|v\|=1}\inf \{\|v-w\| \colon w \in W\}\right\}$$</code> This is I think not <em>quite</em> the same as the one suggested by Ryan Budney, but produces the same topology. This one seems to be the most popular definition for people working in multiplicative ergodic theory (it is in Barreira and Pesin's book, for example).</p> <p>There are some equivalent ways of describing this metric which seem to be less well-known. If we know <em>a priori</em> that $V$ and $W$ have the same dimension, then the maximum in the expression above is always attained by both expressions simultaneously! Hence if we fix a dimension $r$, then the expression <code>$$d(V,W):=\sup_{v \in V,\|v\|=1}\inf\left\{\|v-w\|\colon w \in W\right\}$$</code> is actually a metric for the component of the Grassmannian which consists of all $r$-dimensional subspaces. This does not seem to be very well-known; I actually discovered this by reading Kato's book on perturbation theory, which isn't exactly the first place I'd go to to find out about Grassmannian manifolds...</p> <p>Another way to put a metric on the Grassmannian is as follows. We can identify a subspace $U$ with the unique linear operator of orthogonal projection onto that subspace, and take the metric given by setting the distance between two subspaces to be the operator norm distance between the orthogonal projection operators corresponding to those subspaces. I personally like this approach a great deal, because I think it makes it very obvious that the Grassmannian is compact (well, obvious if you're a functional analyst!). This metric is also, rather pleasantly I think, exactly identical to the first metric I defined above. You can find a proof that the two things are the same in the book on Hilbert spaces by Akhiezer and Glazman.</p> <p>There's a short discussion on this topic in my paper "A rapidly-converging lower bound for the joint spectral radius via multiplicative ergodic theory", which is basically the result of a gentle argument between myself and the referee over how the metric on the Grassmannian should be defined! </p> http://mathoverflow.net/questions/48088/unique-equilibrium-states-for-systems-without-specification/48112#48112 Answer by Ian Morris for Unique equilibrium states for systems without specification Ian Morris 2010-12-02T22:34:13Z 2010-12-03T18:41:30Z <p>Let $T_1 \colon X_1 \to X_1$ be an Anosov diffeomorphism and let $T_2 \colon X_2 \to X_2$ be a uniquely ergodic expansive homeomorphism which is not a periodic orbit. Let $T \colon X_1 \times X_2 \to X_1 \times X_2$ be given by the direct product of the two maps. Clearly $T$ is expansive, and $T$ does not have specification because it has no periodic orbits. The invariant measures of $T$ are precisely the products of the invariant measures of $T_1$ with the unique invariant measure of $T_2$ (right...?). So, calculating the equilibrium state(s) of a H&ouml;lder function defined on $X$ is the same problem as calculating the equilibrium state(s) of the function on $X_1$ defined by integrating $f$ along fibers against the unique invariant measure of $T_2$. The fiberwise integral has to be H&ouml;lder because $f$ is H&ouml;lder, and it follows that $f$ has a unique equilibrium state.</p> <p>Edit: the sentence beginning "The invariant measures of $T$ are precisely..." is probably wrong - see comments below.</p> http://mathoverflow.net/questions/47080/examples-of-transformations-which-are-weak-mixing-but-not-strong-mixing Examples of transformations which are weak-mixing but not strong-mixing Ian Morris 2010-11-23T10:53:24Z 2010-11-24T05:07:48Z <p>I was reminded of this topic by some of the answers to <a href="http://mathoverflow.net/questions/46966" rel="nofollow">this question</a>, where it was noted that "typical" measure-preserving transformations are weak-mixing but not strong-mixing for several senses of "typical". As a result, it occurred to me that I do not know of any very natural, explicit examples of transformations which are weakly but not strongly mixing. So,</p> <blockquote> <p>What are some good examples of measure-preserving transformations which are weak-mixing but not strong-mixing?</p> </blockquote> <p>To clarify "good": I'm particularly interested in examples where it can be proved in a concise and self-contained manner that weak mixing occurs and strong mixing does not, in examples which arise constructively, and in examples which arise directly from a continuous transformation of a compact metric space (as opposed to abstract measure-theoretic constructions).</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/46966/what-do-singular-atomless-invariant-measures-of-times-d-look-like/47013#47013 Answer by Ian Morris for What do singular, atomless invariant measures of $\times d$ look like? Ian Morris 2010-11-22T22:43:14Z 2010-11-22T22:43:14Z <p>To supplement the previous answers (and perhaps further illustrate that there are <em>a lot</em> of invariant measures!) I thought I'd mention what "typical" invariant measures for the $d$-fold expanding map look like. Fix a $d$-fold expanding map $T$, and consider the set $\mathcal{M}$ of all $T$-invariant Borel probability measures on the circle. We can make $\mathcal{M}$ into a compact, metrisable topological space using the weak topology, which is characterised by the fact that a sequence of measures $(\mu_n)$ converges to $\mu$ if and only if the real sequence $(\int f d\mu_n)$ converges to $\int fd\mu$ for every continuous function $f$ from the circle to the reals. We can then ask what <em>typical</em> elements of $\mathcal{M}$ look like in the sense of Baire category.</p> <p>It turns out that the answer is this: there is a dense $G_\delta$ subset of $\mathcal{M}$ in which every measure is fully supported, weak-mixing for $T$, but not strong-mixing for $T$. All of these measures are non-atomic and pairwise mutually singular with respect to one another. So there are enormously many fully supported invariant measures - so many that it's not easy to say anything about them at all which carries with very much generality.</p> <p>This result is basically due to K. R. Parthasarathy in the 1961 paper "On the category of ergodic measures". In that paper it's proved for the two-sided full shift on $d$ symbols (or even infinitely many symbols) but the same proofs go through with little change. </p> http://mathoverflow.net/questions/130572/linear-algebra-span Comment by Ian Morris Ian Morris 2013-05-14T11:50:35Z 2013-05-14T11:50:35Z MathOverflow is intended for research-level questions. This question would be better suited to Mathematics Stack Exchange. http://mathoverflow.net/questions/130295/variational-principle-for-the-entropy/130303#130303 Comment by Ian Morris Ian Morris 2013-05-10T21:37:26Z 2013-05-10T21:37:26Z It's Theorem 8.6 in Walters. http://mathoverflow.net/questions/127799/recurrence-and-transience-of-cocycle-over-a-dynamical-system Comment by Ian Morris Ian Morris 2013-04-18T10:26:01Z 2013-04-18T10:26:01Z Perhaps I am missing something, but how is the inequality $\lim_{n \to \infty} \int \phi_n d\mu \geq \int \liminf_{n \to \infty}\phi_n d\mu$ justified? Fatou's lemma does not work here, for example, because the functions $\phi_n$ might fail to be uniformly bounded below. http://mathoverflow.net/questions/127799/recurrence-and-transience-of-cocycle-over-a-dynamical-system Comment by Ian Morris Ian Morris 2013-04-17T08:31:14Z 2013-04-17T08:31:14Z What is the question? http://mathoverflow.net/questions/124903/liveranis-clt-a-question/125292#125292 Comment by Ian Morris Ian Morris 2013-03-22T19:30:22Z 2013-03-22T19:30:22Z The absolute value of the weak limit of $f_n$ does not have to equal the weak limit of the sequence of absolute values $|f_n|$. For example, in $L^2([0,1])$ take $f_n(x)=sin(nx)$ to obtain $f_n \to 0$ weakly and $|f_n| \to \frac{1}{2}$ weakly. http://mathoverflow.net/questions/124903/liveranis-clt-a-question/125292#125292 Comment by Ian Morris Ian Morris 2013-03-22T19:14:22Z 2013-03-22T19:14:22Z Let $(f_n)$ be a sequence which converges in the weak topology of $L^2$ to $f$, and converges a.e. to $g$. To complete the proof we must show that $f=g$. Let $\delta&gt;0$ and choose a set $E$ with $m(E)&gt;1-\delta$ such that $f$ is bounded on $E$. Since $f_n \to g$ a.e, by Egoroff's Thm we can find $F \subset E$ with $m(F)&gt;1-2\delta$ such that $f_n \to g$ uniformly on $F$. By weak convergence $\int f_n(g-f)\chi_F dm \to \int f(g-f)\chi_F dm$, and by uniform convergence on $F$ also $\int f_n(g-f)\chi_F dm \to \int g(g-f)\chi_F dm$, so $\int_F(g-f)^2=0$ and $f=g$ except on a set of measre $2\delta$. http://mathoverflow.net/questions/124903/liveranis-clt-a-question/125292#125292 Comment by Ian Morris Ian Morris 2013-03-22T18:20:08Z 2013-03-22T18:20:08Z If $(e_n)$ is an orthonormal basis sequence for $L^2$ then its weak limit is zero but it does not converge in the $L^2$ distance. Your argument shows -- correctly I think, if you directly use the definition of weak convergence to justify $E(D_1(\lambda_n I_A)) \to E(D_1I_A)$ instead of attempting to use norm convergence in $L^2$ and $L^1$ -- that $D(\lambda_n)$ has a limit in the weak topology, and that the conditional expectation with respect to $\mathcal{F}_1$ of that limit is zero. It remains only to show that the weak limit really is $D_1$: <a href="http://math.stackexchange.com/questions/160306" rel="nofollow">math.stackexchange.com/questions/160306</a> http://mathoverflow.net/questions/124903/liveranis-clt-a-question/125292#125292 Comment by Ian Morris Ian Morris 2013-03-22T16:59:02Z 2013-03-22T16:59:02Z I am unsure about the details of your application of Alaoglu's theorem: in general a bounded sequence in $L^2$ will not have a convergent subsequence with respect to the $L^2$ distance, only with respect to the weak topology. However, $g \mapsto \mathbb{E}(g\chi_A)$ is a continuous linear functional on $L^2$ so Alaoglu's theorem delivers the result you need and $\mathbb{E}(D_1(\lambda_n)\chi_A)$ does converge to $\mathbb{E}(f\chi_A)$ where $f$ is the limit of the subsequence. I don't recall the details, but it should not be hard to show that the weak limit and a.e. limit agree when both exist. http://mathoverflow.net/questions/125245/silly-question-about-mixing/125268#125268 Comment by Ian Morris Ian Morris 2013-03-22T15:22:13Z 2013-03-22T15:22:13Z Here is a correct modification of Lemma 2.4: if $\mu(T^{-n}A \cap B)$ is eventually nonzero whenever $\mu(A)$ and $\mu(B)$ are both nonzero then $T$ is light mixing. Proof: suppose that $T$ is not light mixing. Choose $A,B$ with $\mu(A),\mu(B)&gt;0$ and $\liminf_{n \to \infty} \mu(T^{-n}\cap B)=0$. Choose a strictly increasing sequence $(n_k)$ such that $\mu(T^{-n_k}A \cap B)&lt;3^{-k}\mu(B)$ for all $k \geq 1$. Let $C:=B \setminus \bigcup_{k=1}^\infty \left(T^{-n_k}A \cap B\right)$. Then $\mu(C)&gt;\frac{1}{2}\mu(B)&gt;0$ and $\mu(T^{-n_k}A \cap C)=0$ for all $k$, a contradiction. http://mathoverflow.net/questions/125245/silly-question-about-mixing/125268#125268 Comment by Ian Morris Ian Morris 2013-03-22T15:13:11Z 2013-03-22T15:13:11Z You're right: if $T$ is not lightly mixing then there is no reason why we should be able to find $E$ such that $\liminf_{n \to \infty}\mu(T^{-n}E\cap E)=0$. I think that the authors err when they state that in order to check light mixing it is sufficient to check the case $A=B$: this is fine for weak, strong and probably mild mixing because in those cases the relevant expressions are linear in $\chi_A$ and $\chi_B$, but lim inf is of course not linear. http://mathoverflow.net/questions/125245/silly-question-about-mixing/125268#125268 Comment by Ian Morris Ian Morris 2013-03-22T14:17:22Z 2013-03-22T14:17:22Z What a fascinating paper! If I'm not wrong, when $T$ is invertible, Lemma 2.4 in that article shows that &#201;tienne's condition (with positive-measure rather than nonempty intersections) is precisely light mixing. http://mathoverflow.net/questions/125245/silly-question-about-mixing Comment by Ian Morris Ian Morris 2013-03-22T10:42:25Z 2013-03-22T10:42:25Z Perhaps you want $\mu(T^{-n}A \cap B)&gt;0$ rather than nonemptiness, since the former is more natural in a probability space. This is certainly not a silly question. In the positive-measure form this condition implies weak mixing by Theorem 4.31 in Furstenberg's book &quot;Recurrence in Ergodic Theory and Combinatorial Number Theory&quot;. In Parry's book &quot;Topics in Ergodic Theory&quot; (p.89) a transformation is discussed which is weak mixing but does not meet this condition. I suspect that the answer to your question is positive but it may not be widely known. http://mathoverflow.net/questions/122764/characterising-ergodicity-of-continuous-maps/122769#122769 Comment by Ian Morris Ian Morris 2013-02-24T16:52:40Z 2013-02-24T16:52:40Z @Julian: Every invariant probability measure of a minimal transformation is fully supported, because otherwise its support would be a nonempty closed invariant proper subset, contradicting minimality. So the two statements are equivalent. http://mathoverflow.net/questions/122764/characterising-ergodicity-of-continuous-maps/122769#122769 Comment by Ian Morris Ian Morris 2013-02-24T12:08:36Z 2013-02-24T12:08:36Z @Julian: this is equivalent to asking for a condition on $X$ such that every minimal transformation on $X$ is uniquely ergodic, i.e. has only one invariant measure. (If a transformation has two distinct invariant measures then a strict linear combination of the two is never ergodic.) Such conditions do exist: finite spaces $X$ have this property, as does the circle (I think) but as Anthony says this is a severly restrictive requirement. The broader stroke of your question seems to be whether ergodicity can be easily characterised using only topological concepts. The answer to this is &quot;No&quot;. http://mathoverflow.net/questions/121227/invariant-measures-for-cellular-automata Comment by Ian Morris Ian Morris 2013-02-08T22:14:13Z 2013-02-08T22:14:13Z If I understand you correctly then the action of this system on the closed subset $\{0,1\}^{\mathbb{N}}$ is simply the shift, so this system admits a host of shift-invariant ergodic measures supported on $\{0,1\}^{\mathbb{N}}$. Or do you want the measure to be fully supported?