User damek davis - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:39:44Z http://mathoverflow.net/feeds/user/11568 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28788/nontrivial-theorems-with-trivial-proofs/65922#65922 Answer by Damek Davis for nontrivial theorems with trivial proofs Damek Davis 2011-05-25T03:17:26Z 2011-05-25T03:17:26Z <p>Chebyshev's inequality is the following:</p> <blockquote> <p>Suppose $X, \mu$ is a measure space, and $f \in L^p(X, \mu)$, then for all $t > 0$ </p> <p>$\mu( {x \in X : |f(x)| \geq t } ) \leq \frac{1}{t^p} \|f\|_{L^p(X, \mu)}^p$. </p> </blockquote> <p>The proof is trivial: </p> <blockquote> <p>Observe that </p> <p>$\mu( {x \in X : |f(x)| \geq t } )t^p = \int_{X} 1_{|f| \geq t}(x)t^p \leq \int_{X} |f|^p = \|f\|_{L^p(X, \mu)}^p$</p> <p>and divide both sides by $t^p$. </p> </blockquote> <p>This is a fundamental inequality in the the study of <a href="http://terrytao.wordpress.com/2009/03/30/245c-notes-1-interpolation-of-lp-spaces/" rel="nofollow">the interpolation of L^p spaces</a>. </p> http://mathoverflow.net/questions/48299/more-open-problems/59997#59997 Answer by Damek Davis for More open problems Damek Davis 2011-03-29T17:39:06Z 2011-03-29T17:39:06Z <p>Igor Shparlinski maintains a large list of open problems in exponential and character sums. You can find it on his website: <a href="http://web.science.mq.edu.au/~igor/" rel="nofollow">http://web.science.mq.edu.au/~igor/</a></p> http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th Is there a category structure one can place on measure spaces so that category-theoretic products exist? Damek Davis 2010-12-14T20:06:32Z 2011-01-09T05:03:24Z <p>The usual category of measure spaces consists of objects $(X, \mathcal{B}_X, \mu_X)$, where $X$ is a space, $\mathcal{B}_X$ is a $\sigma$-algebra on $X$, and $\mu_X$ is a measure on $X$, and measure preserving morphisms $\phi \colon (X, \mathcal{B}_X, \mu_X) \to (Y, \mathcal{B}_Y, \mu_Y)$ such that $\phi_\ast \mu_X(E) = \mu_X(\phi^{-1}(E)) = \mu_Y(E)$ for all $E \in \mathcal{B}_Y$. </p> <p>The category of measurable spaces consists of objects $(X, \mathcal{B}_X)$ and measurable morphisms $\phi \colon (X, \mathcal{B}_X) \to (Y, \mathcal{B}_Y)$. </p> <p>Products exist in the category of measurable spaces. They coincide with the standard product $(X \times Y, \mathcal{B}_X \times \mathcal{B}_Y)$, where $X \times Y$ is the Cartesian product of $X$ and $Y$ and $\mathcal{B}_X \times \mathcal{B}_Y$ is the coarsest $\sigma$-algebra on $X\times Y$ such that the canonical projections $\pi_X \colon X \times Y \to X$ and $\pi_Y \colon X \times Y \to Y$ are measurable. Equivalently, $\mathcal{B}_X \times \mathcal{B}_Y$ is the $\sigma$-algebra generated by the sets $E \times F$ where $E \in \mathcal{B}_X$ and $F \in \mathcal{B}_Y$.</p> <p>However, in the category of measure spaces, products do not exist. The first obstacle is that the canonical projection $\pi_X \colon X \times Y \to X$ may not be measure preserving. A simple example is the product of $(\mathbf{R}, \mathcal{B}[\mathbf{R}], \mu)$ with itself, where $\mathcal{B}[\mathbf{R}]$ is the Borel $\sigma$-algebra on $\mathbf{R}$. In this case, </p> <p><code>$(\pi_{\mathbf{R}})_\ast\mu\times \mu([0,1]) = \mu\times \mu(\pi_\mathbf{R}^{-1}([0,1])) = \mu \times \mu([0,1]\times \mathbf{R}) = \infty \neq 1 = \mu([0,1])$</code>. </p> <p>In addition, there may be multiple measures on $X\times Y$ whose pushforwards on $X$ and $Y$ are $\mu_X$ and $\mu_Y$. <a href="http://damekdavis.wordpress.com/2010/12/13/is-the-product-of-measurable-spaces-the-categorical-product/" rel="nofollow">Terry Tao mentions that</a> from the perspective of probability, this reflects that the distribution of random variables $X$ and $Y$ is not enough to determine the distribution of $(X, Y)$ because $X$ and $Y$ are not necessarily independent. </p> <p>Given that the products in the usual category fail to exist, is it possible to define a new categorical structure on the class of measure spaces such that products do exist?</p> http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th Comment by Damek Davis Damek Davis 2010-12-15T04:05:42Z 2010-12-15T04:05:42Z The counter example is that $\Delta^{-1}(\{(x,x) x\in [0,1]\}) = [0,1]$ has measure $1$ while the diagonal has measure 0. http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th Comment by Damek Davis Damek Davis 2010-12-15T04:05:32Z 2010-12-15T04:05:32Z No Problem :). I added a comment to clarify the meaning of $\mathcal{B}_X \times \mathcal{B}_Y$. If we cannot relax the condition of &quot;measure preserving maps,&quot; then the product of the two spaces $X$ and $Y$ cannot be $X\times Y$ in general. Indeed, this forces the condition that $\pi_X$ and $\pi_Y$ are measurable. This forces the $\sigma$-algebra on $X\times Y$ to contain $\mathcal{B}_X \times \mathcal{B}_Y$. Thus, the counterexample of the diagonal map $\Delta : \mathbf{R} \to \mathbf{R} \times \mathbf{R}$ is available to us, a contradiction. http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th/49473#49473 Comment by Damek Davis Damek Davis 2010-12-15T02:38:19Z 2010-12-15T02:38:19Z Even if we restrict to probability spaces and use the standard Carath&#233;odory construction of product measure, the diagonal map $\Delta \colon [0,1] \to [0,1]\times [0,1]$ is not Borel measure preserving. Indeed, the line $E = \{(x,x) : x \in [0,1]\}$ has measure zero, while $\Delta^{-1}(E) = [0,1]$ has measure one. http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th/49431#49431 Comment by Damek Davis Damek Davis 2010-12-15T01:01:50Z 2010-12-15T01:01:50Z Whoops! I meant to say that $\Delta^{-1}(\{y=x\}) = [0,1]$. This has measure $1$. http://mathoverflow.net/questions/49426/is-there-a-category-structure-one-can-place-on-measure-spaces-so-that-category-th/49431#49431 Comment by Damek Davis Damek Davis 2010-12-15T01:00:33Z 2010-12-15T01:00:33Z What about the line $\{y=x\}$ in $\mathbf{R}^2$ for $x \in [0, 1]$. This has Borel measure zero, but $\mu_R(\Delta^{-1}(\{y=1\})) = \mu_R([0,1]) = 1$