An algebra of "integrals" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:35:49Z http://mathoverflow.net/feeds/question/115743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115743/an-algebra-of-integrals An algebra of "integrals" 36min 2012-12-07T20:23:26Z 2012-12-10T10:38:46Z <p>When discussing divergent integrals with people, I got curious about the following:</p> <p>Is there an $\mathbb{R}$-algebra $A$ together with a map (could be defined on just a subspace)</p> <p>$$\int_0^{\infty}: C^{\infty}(\mathbb{R})--\to A$$</p> <p>which behaves like integration, and is defined even for some function whose integration is divergent in the usual sense? Or,</p> <blockquote> <p>Can we find some universal target of integration which is like the module of Kahler differentials as the universal target of derivation?</p> </blockquote> <p>In other words, is there an $\mathbb{R}$-algebra $A$ together with a map $T: C^{\infty}(\mathbb{R})\to A$ such that (here is a list of plausible properties)</p> <p>$$T(f(x))=\int_0^{\infty}f(x)dx\in\mathbb{R},\text{ if the RHS converge}.$$ $$T(f(x))-T(f(x+a))=\int_0^af(x)dx \text{ for any }f(x).$$ $$T(f(x))=aT(f(ax)) \text{ for } a>0.$$</p> <p>(Edit: I removed the 4th, which is included in the first.)</p> <p>Or for some reasons such an $\mathbb{R}$-algebra could only be $\mathbb{R}$? I tried to construct a free'' algebra of some kind but it is not clear to me what I got. (from the conditions above there are too many generators and relations, and there are even things from "integration by parts" given the last rule, I'm not sure what the quotient of the generators by the relations gives.)</p> <p>(Edit: people asked why $A$ need to be an algebra, I don't have a good reason for that. I just want to see if one can extend the definition of integration so that they land in some vector space with a certain structure. The most naive thing I can think about is that the result of an integral is a certain kind of "number", and we add, subtract, multiply numbers.)</p> http://mathoverflow.net/questions/115743/an-algebra-of-integrals/115789#115789 Answer by S. Carnahan for An algebra of "integrals" S. Carnahan 2012-12-08T10:29:01Z 2012-12-10T10:38:46Z <p>I don't see why you want $A$ to be an algebra, since the integral of 1 doesn't seem like a reasonable unit. Did you want some compatibility with higher dimensional integrals using the Fubini theorem? Otherwise, if you follow Kähler's lead, it seems more natural to expect a real (or complex) vector space.</p> <p>Let $C^\infty(\mathbb{R})_{int}$ denote the subspace of $C^\infty(\mathbb{R})$ whose elements are integrable on $[0,\infty)$, and let $C^\infty(\mathbb{R})_{int}^0$ denote the codimension one subspace of functions whose integral is zero. Here's a rephrasing of the desired properties of $A$ and $T$:</p> <ol> <li><p>Linearity of $T$.</p></li> <li><p>The restriction of $T$ to $C^\infty(\mathbb{R})_{int}$ lands in a distinguished subspace $\mathbb{R} \subset A$, and is given by ordinary integration.</p></li> <li><p>Good behavior under the action of the group $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$ generated by translations and orientation-preserving dilations.</p></li> </ol> <p>[Edit:] Let $X$ is a space of smooth functions closed under addition by $C^\infty(\mathbb{R})_{int}$, such that $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$ acts freely on the quotient vector space $X/C^\infty(\mathbb{R})_{int}$. If a universal target $A$ for integration existed, then $X/C^\infty(\mathbb{R})_{int}^0$ should admit an injection to $A$, because your list of conditions specifies no further relations. The problem (as pointed out by Tao) is that there are lots of smooth functions with nontrivial stabilizer in $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$.</p> <p>I think a common method to removing such a difficulty is to ignore the requirement that integration be $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$-equivariant. Then your universal space is just $C^\infty(\mathbb{R})/C^\infty(\mathbb{R})_{int}^0$.</p> http://mathoverflow.net/questions/115743/an-algebra-of-integrals/115790#115790 Answer by Igor Khavkine for An algebra of "integrals" Igor Khavkine 2012-12-08T11:05:41Z 2012-12-08T16:47:58Z <p>Why make $A$ an algebra? Integration is fundamentally a linear operation that has at best a complicated relationship with multiplication on $C^\infty(\mathbb{R})$. And I see that Scott Carnahan has just made the same point in another answer... Scott's the kernel of what I was going to suggest as well: asymptotic growth classes. Let me expand on that.</p> <p>Let $I_+ \subset C^\infty(\mathbb{R})$ be the ideal of functions that vanish on some neighborhood of $+\infty$ (for definiteness, say, on at least one interval of the form $[a,+\infty)$ with $a\in\mathbb{R}$). Let $A_+ = C^\infty(\mathbb{R})/I_+$. The quotient $A_+$ is an $\mathbb{R}$-algebra whose elements capture the rates of asymptotic growth at $+\infty$. Let $1_+$ be the image of the constant function $1\in C^\infty(\mathbb{R})$ under the quotient map. Define $I_-$, $A_-$ and $1_-$ in the same way, by replacing $+\infty$ with $-\infty$.</p> <p>Now, let $B=A_+\oplus A_-$ and $N\subset B$ be the linear subspace spanned by the element $1_+\oplus 1_-$. And finally let $A = B/N$, where we are just taking the quotient of linear spaces (the ring property of $B$ ceases to be of importance).</p> <p>Any smooth function $f\in C^\infty(\mathbb{R})$ has an indefinite integral $f_a(x) = \int_a^x f(y) dy$ that is also in $C^\infty(\mathbb{R})$. Applying the quotient maps above, we end up with an image $[f_a]$ of $f_a$ in the linear space $A$. The fact that the quotient images of constant functions give zero shows that the images indefinite integrals with different base points (say $f_a$ and $f_b$) coincide. I think that letting $T(f) = [f_a]$ will satisfy all the properties that you wanted of an "integration" map.</p> http://mathoverflow.net/questions/115743/an-algebra-of-integrals/115837#115837 Answer by Will Sawin for An algebra of "integrals" Will Sawin 2012-12-08T20:03:18Z 2012-12-08T20:12:45Z <p>EDIT: Let $g(x)$ be any function that is smooth at $0$, meaning all its derivatives vanish at $0$. Let $f(x)$ satisfy $f(x)=g(x)+f(x-1)$ for $x>0$ and $f(x)=0$ for $x\leq 0$. Clearly $f$ is smooth. Then</p> <p>$\int_0^{\infty} g(x) = \int_0^\infty f(x) - \int_0^\infty f(x-1) = \int_0^\infty f(x) - \int_0^\infty f(x)=0$</p> <p>So the second axiom makes any function which is smooth at $0$ vanish. This contradicts the first axiom.</p> <hr> <p>You can throw out all the axioms but the first, add an obvious algebra structure, and still run into trouble.</p> <p>Using integration by parts, we can see that $\int u dv + \int v du = [uv]_0^{\infty}$. If $u(0)=v(0)=0$, then this is equal to $[u]_0^\infty [v]_0^\infty$. </p> <p>By $[u]_0^\infty$, I mean $\int_0^\infty du$.</p> <p>If you accept that equality, and you want the integral of anything that is actually integrable with integral $0$ to be $0$, then you've got a problem. Take $v$ to be a function that decreases to $0$ at $\infty$ but is never $0$, and take $u=f/v$ for any function $f$ (that vanishes to second order at $0$). Then</p> <p><code>$\int df= \int udv+\int vdu= [u]_0^{\infty} [v]_{0}^\infty = [u]_0^\infty 0 = 0$</code></p> <p>Since the local conditions on $f$ at $0$ clearly don't matter, the integral of any function is zero.</p> <p>It's clear why this happens. Using $\int_0^\infty f(x) dx = \lim_{y\to \infty} \int_0^y f(x) dx$, we see that integration is exactly the same problem as taking limits. The only way to take the limit of an arbitrary function is an ultrafilter. But ultrafilters usually take decreasing functions to (invertible) infinitesimals, not $0$.</p> <p>So this is a good explanation of why you're forced to give up the algebra structure, I think - it prevents you from sending functions that integrate to $0$ to $0$.</p> http://mathoverflow.net/questions/115743/an-algebra-of-integrals/115851#115851 Answer by Terry Tao for An algebra of "integrals" Terry Tao 2012-12-08T22:48:48Z 2012-12-08T22:59:30Z <p>Property (2) gives $T(1) = T(1)+a$ for any real $a$, which is not solvable in any real algebra (or vector space) $A$. Property (3) leads to a similar issue as it implies that $T(1) = aT(1)$ for all $a>0$. </p> <p>Note that many common ways of evaluating divergent sums and integrals (e.g. zeta function regularisation) do not actually obey (2) or (3). For instance, the famous identity $1 + 2 + 3 + \ldots = -1/12$, which is valid if the LHS is summed using zeta function regularisation, is inconsistent with basic axioms such as (2), as discussed in <a href="http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/" rel="nofollow">this blog post of mine</a>. Also, none of these methods are able to sum <em>all</em> divergent series (or integrate <em>all</em> non absolutely integrable functions). In view of this, I doubt that an axiomatic approach that assumes that all integrals can be integrated is the most natural way to proceed here.</p> <p>ADDED LATER: Using enough abstract nonsense, one can integrate arbitrary functions, but in a rather useless way. For instance, using nonstandard analysis, one can map $f \in C^\infty({\bf R})$ to the nonstandard real number $\int_0^N f(x)\ dx \in {}^* {\bf R}$ for some fixed unbounded real number $N$, and this will be a perfectly well defined additive homomorphism. If one quotients out the infinitesimals $o({\bf R}) := \{ x \in {}^* {\bf R}: x = o(1) \}$ from ${}^* {\bf R}$, one in fact gets a real-linear map that obeys property (1) (if one identifies ${\bf R}$ with a subspace of ${}^* {\bf R}/o({\bf R})$ in the usual manner), but not (2) or (3). But I'm not sure one can do anything particularly interesting with this construction.</p>