Can distribution theory be developed Riemann-free? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:42:16Z http://mathoverflow.net/feeds/question/72450 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free Can distribution theory be developed Riemann-free? Paul Siegel 2011-08-09T07:41:56Z 2013-02-05T02:07:09Z <p>I imagine most people who frequent MO have been indoctrinated into the point of view that the Riemann integral can be safely discarded once one has taken the time to develop the Lebesgue integral. After all the two integrals agree more or less whenever they are both defined, and the Lebesgue theory is well known to be more robust and flexible in a lot of important ways.</p> <p>However, I have recently encountered an apparent counter-example to the extreme view (which perhaps nobody actually holds) that the Riemann integral is entirely dispensable as a technical tool. The context is the theory of distributions. It is not uncommon that when one wants to generalize an operation from test functions to distributions that there are two natural choices: the operation can either be defined "directly" or by specifying how it pairs with test functions. Here are two basic examples:</p> <ul> <li>The first example involves the convolution of a distribution $F$ with a test function $\psi$. The direct definition is given by $F \ast \psi(x) = \langle F, \psi_x \rangle$ where $\psi_x(y) = \psi(x-y)$. The definition by pairing stipulates that for any test function $\phi$, $\langle F \ast \psi, \phi \rangle = \langle F, \phi \ast \psi_0 \rangle$.</li> <li>The second example involves the Fourier transform of a (tempered) distribution $F$. The direct definition is given by $\hat{F}(\xi) = \langle F, e_\xi \rangle$ where $e_\xi(x) = e^{2 \pi i \xi x}$. The definition by pairing just sets $\langle \hat{F}, \psi \rangle = \langle F, \hat{\psi} \rangle$ for any appropriate test function $\psi$.</li> </ul> <p>In both of these examples, and others like them, all of the authors that I have consulted (including Folland and Taylor) prove that the direct definition agrees with the definition by pairing by carrying out a calculation with Riemann sums.</p> <p>So I am left wondering if there decent proofs of these results for ordinary Lebesgue-abiding citizens. This question is a little problematic since the Lebesgue integral and the Riemann integral agree on the relevant space of functions, but if there isn't a good affirmative answer then it seems to me that there should be a convincing explanation why measure theoretic tools aren't strong enough to make the argument work.</p> http://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free/72534#72534 Answer by Phil Isett for Can distribution theory be developed Riemann-free? Phil Isett 2011-08-10T01:41:01Z 2011-08-10T01:41:01Z <p>Regarding the first example: there is essentially no way to get around the "Riemann integration". Often when you use it (for example, to characterize monotonic functions) the distributions in question are measures. In this case, you can use Fubini's theorem to interchange the order of integration. Otherwise the statement you're trying to prove is:</p> <p>$\int \int u(x) \phi(y - x) dx \psi(y) dy = \int u(x) (\int \phi(y) \psi(y + x) dy) dx$</p> <p>where the $dx$ integral is to be viewed in the sense of distributions. A priori from the definition of a distribution, this formula is only clear when $\phi$ is a delta function (in which case the convolution is just a translation) or a linear combination thereof. So to prove the general case you will have to use the continuity in the definition of a distribution to pass from the limit by approximating $\phi(x) dx$ with point masses. This is essentially an exercise which is often done in Riemann integration, although you have to keep track of the error and make sure the convergence is in $C^k$. </p> <p>In Rudin's book, this discrete approximation is actually how he constructs the Lebesgue measure from scratch, and it's interesting that it actually even works in the measure case because you are bypassing the use of Fubini's theorem. From that point of view it isn't completely a "Riemann integral" approach.</p> http://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free/120793#120793 Answer by Sönke Hansen for Can distribution theory be developed Riemann-free? Sönke Hansen 2013-02-04T19:07:37Z 2013-02-04T19:07:37Z <p>The equations of the examples follow by interchanging integrals and duality brackets as follows: <code>$$\langle u_y,\int \psi(y,x)dx\rangle=\int\langle u_y,\psi(y,x)dx\rangle.$$</code> Since <code>$u$</code> is a continuous linear functional, this Fubini-type formula follows if the integral converges in the space of test functions; more accurately, if the Riemann sums converge in the topology of the space of test functions in the $y$ variable. To actually prove the convergence is somewhat tedious but straightforward; in a course on distribution theory one has to do this in sufficient detail. In case $u$ is a (locally) integrable function, this argument proves, for the particular case at hand, Fubini's formula via distribution theory.</p> <p>I do not see an issue of Riemann vs. Lebesgue integration here. For fixed $y$ the integral is just a Riemann and/or Lebesgue integral of a test function. </p> http://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free/120823#120823 Answer by Peter Michor for Can distribution theory be developed Riemann-free? Peter Michor 2013-02-05T02:07:09Z 2013-02-05T02:07:09Z <p>An instance where the Riemann integral cannot be replaced by the Lebesque integral is the following theorem: A sequence $x_i$ in the interval $[0,1]$ is equidistributed if and only if for each Riemann integrable function $f$ we have $$\lim_{n\to\infty}\frac1n\sum_{i=1}^n f(x_i) = \int_0^1f(x)dx.$$</p>