User mark peletier - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:23:43Z http://mathoverflow.net/feeds/user/16530 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70611/reference-for-proof-that-c-b-rba Reference for proof that $C_b^* = rba$ Mark Peletier 2011-07-18T11:34:51Z 2011-08-11T14:31:33Z <p>The following theorem seems to have folk status:</p> <p>The topological dual of the space $C_b(X)$ of bounded continuous functions on a topological space $X$ is isomorphic to the space $rba(X)$ of finite, regular, finitely additive Borel set functions.</p> <p>This fact is often mentioned (for instance in the answer to <a href="http://mathoverflow.net/questions/44183/dual-of-bounded-uniformly-continuous-functions" rel="nofollow">http://mathoverflow.net/questions/44183/dual-of-bounded-uniformly-continuous-functions</a>) but I'm having great difficulty actually finding a reference. Often Dunford &amp; Schwartz is mentioned as a reference; D&amp;S defines $rba$, but doesn't prove the connection to the dual of $C_b$. <a href="http://www.jstor.org/pss/1989829" rel="nofollow">Hildebrandt 1934</a> proves a characterization in terms of limits of Stieltjes integrals, but that is still some steps away from the characterization above. I haven't been able to find anything coming closer than this.</p> <p>Does anyone know of a real proof of this statement? Am I maybe overlooking a very simple proof?</p> http://mathoverflow.net/questions/69380/probabilistic-solution-of-the-porous-medium-equation/70620#70620 Answer by Mark Peletier for Probabilistic Solution of the Porous Medium Equation Mark Peletier 2011-07-18T13:28:53Z 2011-07-18T13:41:58Z <p>The processes described by Andre work by having the interaction act at the level of the mobility of the particles. </p> <p>There are other ways, too. One is in the work of <a href="http://sfb611.iam.uni-bonn.de/uploads/232-komplett.pdf" rel="nofollow">Philipowski</a> (see also <a href="http://alea.impa.br/articles/v4/04-09.pdf" rel="nofollow">Figalli &amp; Philipowski</a>). Here the idea is to take interactions of potential type, i.e. for instance</p> <p>$dX^i = -\sum_{j\not=i} \nabla W_\epsilon(X^i-X^j)\, dt + \delta \, dB^i.$</p> <p>The parameter $\epsilon$ is the spatial range of $W$, and in the limit $\epsilon\to0$ the interaction becomes purely local, and leads to a nonlinear diffusion term. If one also lets $\delta\to0$, then the purely Brownian contribution also vanishes. Only the nonlinear diffusion is then left. </p> http://mathoverflow.net/questions/60291/euler-lagrange-gradient-descent-heat-equation-and-image-denoising/70618#70618 Answer by Mark Peletier for Euler-Lagrange, Gradient Descent, Heat Equation and Image Denoising Mark Peletier 2011-07-18T13:18:34Z 2011-07-18T13:18:34Z <p>The equation</p> <p>$u_t = u_{xx} + u_{yy}$</p> <p>is a gradient flow, or gradient descent, in the following sense. You should think of the equation as being placed in the space $L^2$. The Fréchèt derivative of the functional $E$ is the linear mapping</p> <p>$\displaystyle E'(u): v \mapsto -2\iint \nabla u\cdot \nabla v = -2\iint v(u_{xx}+u_{yy}),$</p> <p>where for simplicity I'm assuming that the boundary conditions don't give rise to boundary terms in the partial integration. The second version, after partial integration, is relevant because it's written in the form of an $L^2$ inner product, allowing us to write the Fréchèt derivative as</p> <p><code>$E'(u)\cdot v = (-2(u_{xx}+u_{yy}),v)_{L^2} =: (\mathrm{grad}\ E(u),v)_{L^2}$</code></p> <p>The $L^2$-gradient flow of $E$ is then the equation</p> <p>$u_t = -\mathrm{grad}\ E(u) = 2(u_{xx}+u_{yy}).$</p> http://mathoverflow.net/questions/70611/reference-for-proof-that-c-b-rba Comment by Mark Peletier Mark Peletier 2011-08-14T18:14:26Z 2011-08-14T18:14:26Z Thanks @Theo and @Gerard - I obviously overlooked Theorem IV.6.2, possibly because of the different notation. Thanks for pointing that out! http://mathoverflow.net/questions/57386/is-the-derivative-of-a-lipschitz-function-better-than-l-infty/57418#57418 Comment by Mark Peletier Mark Peletier 2011-07-22T12:47:46Z 2011-07-22T12:47:46Z Of course, in higher dimensions the gradient can not be just any vector-valued $L^\infty$-function $f = (f_i)$, since it satisfies the distributional identity $\dfrac{\partial f_i}{\partial x_j} = \dfrac{\partial f_j}{\partial x_i}.$ So you don't get the whole of $L^\infty$, only those functions that satisfy this identity. http://mathoverflow.net/questions/70917/entropy-of-a-measure/70932#70932 Comment by Mark Peletier Mark Peletier 2011-07-22T12:30:49Z 2011-07-22T12:30:49Z Tapio, why do you want the entropy of your finitely additive translation-invariant measure to be finite? In the limit as $n\to\infty$ the Shannon entropy of the uniform measure on $[n]$ converges to infinity. Wouldn't you therefore expect that such a `uniform' measure on $\mathbb N$ <i>should</i> have infinite entropy?